What is the kinetic energy of a 30 kg dog that is running at a speed of 8.6 m/s (about 19 mi/h)?

Answer:

-67 m/s

Explanation:

We are given that

Mass of ball,m=0.4 kg

Initial speed,u=14 m/s

Impulse,I=-32.4 N-s

Time,t=27 ms=\(27\times 10^{-3} s\)

We have to find the mass's velocity in the x- direction.

We know that

\(Impulse=mv-mu\)

Substitute the values

\(-32.4=0.4v-0.4(14)\)

\(-32.4+0.4(14)=0.4 v\)

\(-26.8=0.4v\)

\(v=\frac{-26.8}{0.4}=-67m/s\)

Answer:

Explanation:

Unknown fork frequency is either

335 + 5.3 = 340.3 Hz

or

335 - 5.3 = 329.7 Hz

After we modify the known fork, the unknown fork frequency equation becomes either

(335 - x) + 8 = 340.3

(335 - x)  = 332.3

x = 2.7 Hz

or

(335 - x) + 8 = 329.7

(335 - x) = 321.7

x = 13.3 Hz

IF the unknown fork frequency was 340.3 Hz,

THEN the 335 Hz fork was detuned to 335 - 2.7 = 332.3 Hz

IF the unknown fork frequency was 329.7 Hz,

THEN the 335 Hz fork was detuned to 335 - 13.3 = 321.7 Hz

The formulated sentences for each will be:

A: If length is 5cm, width 2cm, height 4cm, volume:  40 cubic centimeters.

B: If the length is 3cm, width 4cm, height 2cm, then Volume is: 24 cubic centimeters.

C:  If the length is 6cm, width 2cm, height 3cm, the  volume is 36 cubic centimeters.

D:  If the length is 3cm, width 2cm, height 4cm, the  volume 24 cubic centimeters.

To calculate the volume of a rectangular prism using multiplication, you need to multiply the length, width, and height of the prism. The multiplication sentence for this is:

Volume = length (m) x width (m) x height (m)

For a. The multiplication sentence for calculating the volume of rectangular prism C is:

b. The multiplication sentence for calculating the volume of rectangular prism B is:

c. The multiplication sentence for calculating the volume of rectangular prism C is:

Lastly, for d. The multiplication sentence for calculating the volume of rectangular prism D is:

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(a) The workdone by the car is 4.5 J (b) The potential energy of the cart is 5 J (c) The kinetic energy of cart is 5J.

Work is said to be done when ever a force moves a body through a certain distance.

(a) To calculate the workdone by the car, we use the formula below

Formula:

Where:

Substitute these values into equation 1

(b) To calculate the potential energy the cart have, we use the formula below

Threrefore the potential energy of the cart is 5 J.

(c) Kinetic energy of the cart = work done by the cart = 5 J

Hence, (a) The workdone is 4.5 J (b) The potential energy is 5 J (c) The kinetic energy is 5J.

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After 42 minutes, there are 203 x \(10^{20\) atoms remaining of the radioactive substance.

To determine the number of atoms remaining after a certain amount of time has passed, we can use the formula for radioactive decay:

N(t) = N0 * \((1/2)^{(t / T)\),

where:

N(t) = number of atoms remaining at time t,

N0 = initial number of atoms (812 x  \(10^{20\) atoms),

T = half-life of the substance (21 minutes), and

t = time that has passed.

Let's calculate the number of atoms remaining after a given time.

Suppose the time passed is t = 42 minutes (twice the half-life).

N(t) = 812 x  \(10^{20\) * \((1/2)^{(42 / 21)\)

N(t) = 812 x  \(10^{20\) * \((1/2)^2\)

N(t) = 812 x  \(10^{20\) * 1/4

N(t) = 203 x  \(10^{20\) atoms.

So, after 42 minutes, there are approximately 203 x \(10^{20\) atoms remaining of the radioactive substance.

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Answer:

A. convergent boundary without subduction

Explanation:

The convergent boundary pushes together and makes the plates burst in an upward motion and creates a mountain.

the first one is: it's increased by a factor of 9

the second one: is m3/(kg*s^2)

by what factor does gravitational force between two objects increase if one double in mass and the distance between them decreases by half?: 8

what happens to the gravitational force between two objects that are 15 m apart, when one of them moves 3 m closer?: it increases by a factor of 1 9/16

in thinking about the uniform distribution of a gravitational filed over a sphere, why does the gravitational force have an inverse-square relationship with the distance between objects?: because the surface area of a sphere is proportional to the square if the radius

All these questions are correct I just took the quiz. Hope this helps you and others!

Answer:

1. The force decreases by a factor of 9.

2. 8

3. (N ⋅ m^2) / kg^2

4. It increases by a factor of 1 9/16

5. because the surface area of a sphere is proportional to the square of its radius

Explanation:

The radiation that is least damaging to humans is non-ionizing radiation.

Non-ionizing radiation refers to the type of radiation that does not have enough energy to remove tightly bound electrons from atoms or molecules, thus not causing significant damage to biological tissues.

Examples of non ionizing radiation include radio waves, microwaves, visible light  and low energy ultraviolet (UV) radiation. while excessive exposure to any form of radiation can have adverse effects, non-ionizing radiation is generally considered to be less harmful compared to ionizing radiation, which includes X-rays and gamma rays.

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Answer:

because the friction/movement of the bikes wheels, generates electricity and energy that the light can use to generate electricity and it also does not have a battery on it and relies on the wheels to generate the electricity for the light to work.

Explanation:

The  acceleration, tension of the cord between the 3.5 and 1 block, and the force excreted on the 2 block from the 1 block is 12.6 N.

To solve this problem, we need to apply Newton's second law of motion, which states that the net force acting on an object is equal to the product of its mass and acceleration.

We can start by finding the acceleration of the system. Since the surface is frictionless, there are no horizontal forces acting on the blocks except for the tension in the string. Thus, we can write:

F - T = (m1 + m2 + m3) * a

where F is the force acting on the 3.5 kg block, T is the tension in the string between the 3.5 kg and 1 kg blocks, and a is the acceleration of the system. Substituting the values given in the problem, we get:

46 N - T = (1 kg + 2 kg + 3.5 kg) * a

46 N - T = 6.5 kg * a

Next, we need to find the tension in the string between the 3.5 kg and 1 kg blocks. This tension is the same throughout the string, so we can use the same equation we used to find the acceleration:

T - m1 * g = m1 * a

where g is the acceleration due to gravity (9.8 m/s^2). Substituting the values given in the problem, we get:

T - 1 kg * 9.8 m/s^2 = 1 kg * a

T = 1 kg * (a + 9.8 m/s^2)

Now, we can substitute this expression for T into the first equation we derived to get:

46 N - 1 kg * (a + 9.8 m/s^2) = 6.5 kg * a

Simplifying and solving for a, we get:

a = 2.84 m/s^2

Now that we know the acceleration of the system, we can find the force exerted by the 1 kg block on the 2 kg block. Since there are no horizontal forces acting on the 2 kg block, this force is equal in magnitude and opposite in direction to the force exerted by the 2 kg block on the 1 kg block. Thus, we can write:

F1-2 = -m2 * a

where F1-2 is the force exerted by the 1 kg block on the 2 kg block. Substituting the values given in the problem, we get:

F1-2 = -2 kg * 2.84 m/s^2

F1-2 = -5.68 N

Therefore, the force exerted by the 1 kg block on the 2 kg block is 5.68 N, directed to the left.

Finally, we can use the equation we derived earlier for the tension in the string to find the tension between the 3.5 kg and 1 kg blocks. Substituting the value we found for a, we get:

T = 1 kg * (2.84 m/s^2 + 9.8 m/s^2)

T = 12.6 N

Therefore, the tension in the string between the 3.5 kg and 1 kg blocks is 12.6 N.

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The two evidence are the car's engine gets hot as it burns gasoline are the car remains full of gasoline if its engine stays off.

The car's engine gets hot as it burns gasoline and the car remains full of gasoline if its engine stays off are the two pieces of evidence support the conclusion that the total amount of energy in a close system remains the same because when the gasoline burns, the engine gets hotter and when the car remains full of gasoline if its engine stays off which means no energy leaves the system.

So we can conclude that The two evidence are the car's engine gets hot as it burns gasoline are the car remains full of gasoline if its engine stays off.

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Answer:

If you are asking for the answer with the highest molar mass, then the answer is B.

B. Fe₂O₃

The number of distinct ways of placing four indistinguishable balls into five distinguishable boxes is 20.

To solve this problem, we can use the stars and bars method, which is a combinatorial technique used to count the number of ways to distribute indistinguishable objects into distinguishable containers. In this case, we have four indistinguishable balls and five distinguishable boxes.

The stars and bars method works by representing each ball as a star and using bars to separate the balls into different boxes. For example, if we wanted to distribute two balls into three boxes, we could use the following diagram:

* | * * | *

In this diagram, the first and last bars represent the boundaries of the containers, while the stars represent the balls.

The second bar separates the first two balls from the last ball, indicating that the first two balls are in the first container and the last ball is in the third container.

To distribute four balls into five boxes, we need to use three bars to separate the balls into four groups. We have a total of six spaces to place the bars (including the boundaries), and we need to choose three of them to place the bars.

Therefore, the number of distinct ways to place four indistinguishable balls into five distinguishable boxes is the same as the number of ways to choose three spaces out of six, which is:

6 choose 3 = (6!)/(3!3!) = 20

Therefore, there are 20 distinct ways to place four indistinguishable balls into five distinguishable boxes using the stars and bars method.

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Answer: Skinfold testing

Explanation:

Skinfold testing, is also referred to as calliper testing and it's used to know the body fat percentage. Skinfold testing is an inexpensive, portable, and common way to assess body fat in the fitness industry.

It is typically done with the use of caliper tapes, marker pens which makes it cheap. Skinfold testing isn't usually accurate which is as a result of human errors.

PV=nRT

Here

P=Pressure

V=Volume

n=Molarity

R=universal gas constant

T=Temperature.

LHS

\(\\ \tt\bull\leadsto PV\)

\(\\ \tt\bull\leadsto [ML^2T^{-2}][M^0L^3T^0]\)

\(\\ \tt\bull\leadsto [ML^5T^{-2}]\)

RHS

\(\\ \tt\bull\leadsto nRT\)

\(\\ \tt\bull\leadsto [M^0L^{3}T^0][M^1 L^2 T^{-2}K^{-1}][M^0L^0T^0K^1]\)

\(\\ \tt\bull\leadsto [ML^5T^{-2}]\)

LHS=RHS

hence verified

The answers are (a) |q| = 58.2 cm, M = -1.50, virtual, upright, shrunk; (b) |q| =, M = -, actual, substantially magnified; and (c) |q| = 6.5 cm, M = -1.68, virtual, upright, enlarged.

We can use the magnification formula and the thin lens equation to solve:

Where f is the lens's focal length, d_i is the image distance, and d_o is the object distance, the equation is: 1/f = 1/d_i + 1/d_o.

The formula for magnification is M = -d_i/d_o, where M stands for magnification.

For part (a), the object distance is d_o = -38.8 cm and the focal length is f = -19.4 cm.

1/-19.4 = 1/d_i + 1/-38.8

|d_i| = 58.2 cm

The image is virtual and smaller than the object (since the magnification is less than 1)

For part (b), the object distance is d_o = -19.4 cm, and the focal length is f = -19.4 cm.

1/-19.4 = 1/d_i + 1/-19.4

|d_i| = ∞

The image is at infinity, which means it is a real image and highly magnified (since the object distance is close to the focal length).

For part (c), the object distance is d_o = -9.70 cm, and the focal length is f = -19.4 cm.

1/-19.4 = 1/d_i + 1/-9.70

|d_i| = 6.5 cm

The image is virtual and larger than the object (since the magnification is greater than 1).
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Answer:

I think it does not have a specific shape

Explanation:

Irregular galaxies have no definite shape, which means that the second option is correct. They are definitely not round.

The "spiral galaxy" is the type of galaxy that has arms that extend from the center. These arms look "spiral," which influenced its name.  

They are actually smaller than the other types of galaxies. This makes them prone to collisions.  

The phrase that describes the irregular galaxy is that it does not have a specific shape. The correct option is B.

A galaxy with an undefined shape is an irregular galaxy with a large number of young stars and a large amount of gas and dust that is typically found near larger galaxies.

An irregular galaxy is a catch-all term for any galaxy that does not neatly fit into one of the Hubble classification schemes.

They have no discernible shape or structure and could have formed as a result of collisions, close encounters with other galaxies, or violent internal activity.

They lack beautiful spiral arms, but they do have dark patches of gas and dust. Some irregular galaxies appear to be the result of two galaxies colliding.

Thus, the correct option is B.

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Answer:

Tension in the vertical rope: approximately \(613 \; {\rm N}\).

Tension in the horizontal rope: approximately \(3.74 \times 10^{3}\; {\rm N}\).

Assumption: \(g = 9.81\; {\rm N \cdot kg^{-1}}\).

Explanation:

Since the system is not moving, the tension in the vertical rope would be equal to the weight of the crate:

\(\begin{aligned}\text{weight of crate} &= m\, g \\ &= 62.5\; {\rm kg} \times 9.81\; {\rm N \cdot kg^{-1}} \\ &= 613.25 \; {\rm N}\end{aligned}\).

Note that \(\theta = 57.7^{\circ}\) is the angle between the beam (the lever) and the vertical rope. The torque that this vertical rope exert on the beam would be:

\(\begin{aligned} \tau &= r\, F\, \sin(\theta) \\ &=(7.65\; {\rm m}) \, (613.25\; {\rm N})\, (\sin(57.7^{\circ})) \\ &\approx 3.965 \times 10^{3}\; {\rm N \cdot m} \end{aligned}\).

This torque is in the clockwise direction.

The weight of the beam (\(m = 116\; {\rm kg}\)) would be:

\(\begin{aligned}\text{weight of beam} &= m\, g \\ &= 116 \; {\rm kg} \times 9.81\; {\rm N \cdot kg^{-1}} \\ &= 1.138 \times 10^{3}\; {\rm N}\end{aligned}\).

Note that \(\theta = 57.7^{\circ}\) is also the angle between the beam and the direction of the (downward) gravitational pull on this. Since this beam is uniform, it would appear as if the weight of this beam is applied at the center of this beam (with a distance of \((7.65\; {\rm m}) / 2\) from the pivot.) Thus, the torque gravitational pull exerts on this beam would be:

\(\begin{aligned} \tau &= r\, F\, \sin(\theta) \\ &= \genfrac{(}{)}{}{}{7.65\; {\rm m}}{2} \, (1.138 \times 10^{3}\; {\rm N})\, (\sin(57.7^{\circ})) \\ &\approx 3.679 \times 10^{3}\; {\rm N \cdot m} \end{aligned}\).

This torque is also in the clockwise direction.

The tension in the horizontal rope would need to supply a torque in the counterclockwise direction. The magnitude of that torque would be approximately:

\(\begin{aligned} & 3.965\times 10^{3}\; {\rm N \cdot m} + 3.679\times 10^{3}\; {\rm N \cdot m} \\ \approx \; & 7.645 \times 10^{3}\; {\rm N \cdot m} \end{aligned}\).

Note the angle between the direction of this tension and the beam is \((90^{\circ} - \theta) = 32.3^{\circ}\). This force is applied  \((7.65\; {\rm m}) / 2\) from the pivot. Hence, achieving that torque of \(7.645 \times 10^{3}\; {\rm N \cdot m}\) would require:

\(\begin{aligned} F &= \frac{\tau}{r\, \sin(90^{\circ} - \theta)} \\ &\approx \frac{7.645\times 10^{3}\; {\rm N \cdot m}}{((7.65\; {\rm m}) / 2) \times \sin(32.3^{\circ})} \\ &\approx 3.74 \times 10^{3}\; {\rm N} \end{aligned}\).

Answer: B

Explanation: the energy stored in the bowstring is a result of the elastic nature of the string (k) how far the bow has been stretched (x).

elastic potential energy is the energy stored in any elastic material when stretched.

Explanation:

i expected to use Stefan's law of heat exchange but the value you gave aren't conclusive.

I should say that the temperature of the star should be close to that of the sun because of the similarity in the intensity curves

Answer: I am so sorry i hadn't learned this Yet~!

Explanation:

Answer:

d= 1650 km.

Explanation:

       \(t = \frac{\Delta x}{v} =\frac{1.9m}{346m/s} = 5.5 msec (1)\)

       \(d = v*t = 3e8 m/s*5.5e-3s = 1650 km (2)\)

Answer:

displacement = 55 cm

Explanation:

Initial position = 25 cm

Final position = 80 cm

Displacement = final position-finitial position

Putting values in above formula,

D = 80 cm - 25 cm

D = 55 cm

It means that the displacement of the ant is 55 cm.

The net force on particle q₂ is 1.05 x 10¹³ N, directed towards particle q1.

To find the net force on particle q₂, we need to calculate the force between q₁ and q₂, and the force between q₂ and q₃, and then add these two forces together.

The force between two point charges can be calculated using Coulomb's law, which states that

F = k × q₁× q₂ / r²

where F is the force, k is Coulomb's constant (9.0 x 10⁹ N m² / C²), q₁ and q₂ are the magnitudes of the charges, and r is the distance between the charges.

The force between q₁ and q₂ is

F₁ = k × q₁ × q₂ / r₁²

where r₁ = 0.10 m.

Substituting the values, we get

F₁ = 9.0 x 10⁹ N m² / C²  × 8.0 C  × 3.5 C / (0.10 m)² = 2.52 x 10¹³ N

The force between q₂ and q₃ is

F₂ = k  × q₂  × q₃ / r₂²

where r₂ = 0.15 m.

Substituting the values, we get

F₂ = 9.0 x 10⁹ N m² / C²  × 3.5 C  × (-3.5 C) / (0.15 m)² = -1.47 x 10¹³ N

Note that the negative sign indicates that the force is attractive, since q₂ and q₃ have opposite signs.

The net force on q₂ is the vector sum of F₁ and F₂

\(F_{net}\) = F₁ + F₂ = 2.52 x 10¹³ N - 1.47 x 10¹³ N = 1.05 x 10³ N

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Answer:

75 cm/second.

Explanation:

Formula:

Walking speed = stride length / time per step

Walking speed = 90cm/time per step

                         = 90cm/1.2 seconds (a common estimate time per step)

                         = 75cm/second.

Answer:

equator

Explanation:

in south & north pole you could have 20+ hours daylight or night, everyday!

Answer:

2.687×1025 particles/m3

Astronomers can calculate the pair's size by observing how they tug on one another, and from that they can extrapolate other properties like temperature and radius. All solitary main sequence stars in the universe share these characteristics.

Any binary system has elliptical orbits for both stars around the center of mass (COM). The semi-major axis cuts the long axis of the elliptical in half. A specific kind of ellipse with a radius equal to its semi-major axis is a circle.

The relationship between the orbital period and orbital radius is described by Kepler's third law, sometimes referred to as the law of periods. This law states that the square of an object's orbital period is directly proportional to the cube of its orbital radius, or vice versa.

Therefore, In this instance, the V cube equals two, 3.14, 6.67, and 10 days. To part 30 in 27.45 split by 1.0 in 2-4 in two 3600, power -11 in two, two into 10 days. After solving, we see that v cubed is 42.49 into 10 raised to power 15 m/s, meaning that the value of V is 3.4 into 10 raised to power 5 m/s.

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a) A = 8.0 m South

Since the vector is directly along the South direction, there is no x component.

x component: 0 m

y component: -8.0 m (negative because it's southward)

Component form: A = (0, -8.0)

b) B = -15.0 m at 30° East of North

To find the components, we can use the following relationships:

x component: B_x = B * sin(θ)

y component: B_y = B * cos(θ)

B_x = -15.0 * sin(30°) = -15.0 * 0.5 = -7.5 m

B_y = -15.0 * cos(30°) = -15.0 * (sqrt(3)/2) ≈ -12.99 m

Component form: B ≈ (-7.5, -12.99)

c) C = 12.0 m at 25° South of West

x component: C_x = -C * cos(θ) (negative because it's westward)

y component: C_y = -C * sin(θ) (negative because it's southward)

C_x = -12.0 * cos(25°) ≈ -10.85 m

C_y = -12.0 * sin(25°) ≈ -5.16 m

Component form: C ≈ (-10.85, -5.16)

d) D = 10.0 m at 53° West of North

x component: D_x = -D * sin(θ) (negative because it's westward)

y component: D_y = D * cos(θ)

D_x = -10.0 * sin(53°) ≈ -8.0 m

D_y = 10.0 * cos(53°) ≈ 6.0 m

Component form: D ≈ (-8.0, 6.0)

The pushing force she exerts on the sled is equal to the frictional force the ground exerts on the sled.

In the video, the girl pushed her brother on the sled at a constant velocity, which means the forces acting on the sled must have balanced. Therefore, the pushing force the girl exerted must have been equal to the frictional force that the ground exerted on the sled.

This is because, according to Newton's Third Law of Motion, for every action there is an equal and opposite reaction. In this case, the action is the girl pushing the sled, and the reaction is the frictional force exerted by the ground on the sled. Since the sled is moving at a constant velocity, this means that the forces acting on it are balanced, and therefore the pushing force and the frictional force must be equal.

In equation form, this can be represented as:
F_push = F_friction

Where F_push is the pushing force exerted by the girl, and F_friction is the frictional force exerted by the ground.

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