A block is given an initial speed of 3m/s up a 25° incline coefficient of friction= 0.12
A) how far up the plane will it go?
B) how much time elapses before it returns to it's starting point?.

Answer:

"the minimum frequency of radiation that will produce a photoelectric effect."

Explanation:

That answer was derived from gogle cuz my explanations was harder to explain but good luck

Answer:

5

Explanation:

The peak wavelength for peak intensity of 3000 K is 967 nm and peak wavelength for peak intensity of 50,000K will be 58nm.

Power is the amount of energy which is transferred or converted per unit of time taken. In the International System of Units, the unit of power is the watt (W), which is equal to one joule per second.

Stefan-Boltzmann constant = 5.67 × 10⁻⁸ Wm²

Peak intensity = 2000 K

P = (5.67 × 10⁻⁸) × (3000)⁴

P = 4.59 × 10⁶

P = 459 2700 watts of power per square meter.

Peak intensity = 3000 K

Using Wein law,

λ max = 0.29/ 7

λ max = 0.29/ 3000 = 9.67 × 10⁻⁵ = 967 nm

Same method as above will be used for 50,000 K

P = 3.56 × 10¹¹ W/m²

P = 356.00.000000000 W/m²

λ (Peak) = 58nm

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The magnitude of the force acting at point O is determined as 240 N.

The magnitude of the force at point O is calculated by applying the principles of moment as shown below.

sum of the clockwise moment = sum of the anticlockwise moment

F₀(2 m + 2m + 2m) = 260 N (2m + 2 m ) +  200 N ( 2 m )

where;

6F₀ = 260 (4) + 200(2)

6F₀ = 1,440

F₀ = 1440 / 6

F₀ = 240 N

Thus, the magnitude of the force acting at point O is determined as 240 N.

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B. The resistance R of the circuit is determined as 2.0 ohms.

Ohm's Law states that the current flowing in a circuit is directly proportional to the applied potential difference and inversely proportional to the resistance.

V = IR

R = V/I

The slope of the given graph is the resistance of the circuit.

Choosing point (0 A, 0 V) and (4.5 A, 9 V)

R = ΔV/ΔI

R = (9 - 0 ) / (4.5 - 0)

R = 9/4.5

R = 2.0 ohms

Thus, the resistance R of the circuit is determined as 2.0 ohms.

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Answer:

19

Explanation:

I legit did this and it took 19.

Answer:

The north and south pole of a solenoid depends on two factors. One, the direction of the current flow and two, the direction of the winding (clockwise or counter-clockwise). ... If clockwise in relation to the positive wire then is the south pole, if anti-clockwise then is the north pole.

Explanation:

The north and south pole of a solenoid depends on two factors. One, the direction of the current flow and two, the direction of the winding (clockwise or counter-clockwise). ... If clockwise in relation to the positive wire then is the south pole, if anti-clockwise then is the north pole.

The interval between the first and second echo is 7 seconds. This means that after the initial sound wave reaches the first cliff, it takes a total of 7 seconds for the sound to travel to the second cliff and then return to the student as the second echo.

To determine the interval between the first and second echo, we need to consider the time it takes for sound to travel from the student to the first cliff, and then from the first cliff to the second cliff, and finally back to the student.

Let's break down the distances and calculate the time for each part of the journey:

Distance from the student to the first cliff: 330 meters

Time taken: t₁ = distance / speed = 330 m / 330 m/s = 1 second

Distance from the first cliff to the second cliff: 990 meters

Time taken: t₂ = distance / speed = 990 m / 330 m/s = 3 seconds

Distance from the second cliff back to the student: 990 meters

Time taken: t₃ = distance / speed = 990 m / 330 m/s = 3 seconds

Now, we can calculate the total interval between the first and second echo by adding up the individual times:

Interval between first and second echo = t₁ + t₂ + t₃ = 1 s + 3 s + 3 s = 7 seconds

Therefore, the interval between the first and second echo is 7 seconds. This means that after the initial sound wave reaches the first cliff, it takes a total of 7 seconds for the sound to travel to the second cliff and then return to the student as the second echo.

It's important to note that this calculation assumes a straight path for the sound waves and neglects factors such as air temperature and wind that can affect the speed of sound. Additionally, it assumes perfect reflection of sound waves off the cliffs, which may not be the case in real-world scenarios.

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Note the complete questions is:

A student stands 330m from a tall cliff which is 990m from another tall cliff. If the speed of sound between the cliffs is 330m/s.What is the interval between the first and second echo?

The car is moving at approximately 12.5 meters per second.

The kinetic energy (KE) of an object can be calculated using the formula:

KE = 1/2 * m * \(v^2\)

where

KE = kinetic energy,

m =Mass of the object, and

v = velocity.

In this case, we are given the mass (m) of the car as 1600 kg and the kinetic energy (KE) as 125,000 J. To find the velocity .

Substituting the  values , we have:

125,000 J = 1/2 * 1600 kg *\(v^2\)

Now, we can solve for v by rearranging the equation:

\(v^2\) = (2 * 125,000 J) / 1600 kg

\(v^2\) = 156.25 \(m^2/s^2\)

Taking the square root, we find:

v = √156.25\(m^2/s^2\)

v ≈ 12.5 m/s

Therefore, the car is moving at approximately 12.5 meters per second.

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Explanation:

One idea would be to investigate the correlation between your pulse pressure and your pulse rate.  To do this, you'll need a blood pressure monitor.

First, measure your resting pressure and rate.  Then exercise for 30 seconds.  Measure your new blood pressure and pulse rate.  Wait for your pressure and rate to return to normal, then repeat the trial for 1 minute, 1.5 minutes, 2 minutes, etc.

List the results in a table.  This should include the amount of exercise time, your pulse rate, your systolic pressure (the high number, which is your blood pressure during contraction of your heart muscle), and your diastolic pressure (the low number, which is your blood pressure between heartbeats).  Calculate your pulse pressure (systolic minus diastolic) for each trial.  Graph the pulse pressure on the x-axis, and your pulse rate (beats per minute) on the y-axis.

What do you hypothesize will be the shape of the graph?  Consider Bernoulli's formula, which relates fluid pressure and flow.  How close do the results match your hypothesis?  What might explain any differences?

The coefficient of linear expansion, given that the length of the pipe increased by 1.5 cm is 1.67×10¯⁵ /°F

From the question given above, the following data were obtained

The coefficient of linear expansion can be obtained as illustrated below:

α = ∆L / L₁∆T

α = 0.015 / (10 × 90)

α = 0.015 / 900

α = 1.67×10¯⁵ /°F

Thus, we can conclude that the coefficient of linear expansion is 1.67×10¯⁵ /°F

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The lithosphere is the solid layer of Earth, the hydrosphere is the water-based part, and the atmosphere is the air, they work together to shape the biosphere.

The lithosphere is the soil part that composes the Earth's planet (i.e., the soil), the hydrosphere refers to oceans and all water bodies, while the atmosphere is the layer of air. These three layers interact with biotic factors to shape the biosphere.

In conclusion, the lithosphere is the solid layer of Earth, the hydrosphere is the water-based part, and the atmosphere is the air, they work together to shape the biosphere.

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Answer:

yes bc they are falling :]

Explanation:

The absolute pressure at an ocean depth of 1.0 x 10^3 m is 1.002 x 10^8 Pa.

Hydrostatic pressure is the pressure that a fluid exerts on a surface due to the weight of the fluid above it. It is the result of the force of gravity acting on a column of fluid, and it is directly proportional to the height of the column of fluid and the density of the fluid.

The absolute pressure at an ocean depth of 1.0 x 10^3 m can be calculated using the hydrostatic pressure equation:

P = ρgh + Po

where:

P is the absolute pressure at the given depth

ρ is the density of the water

g is the acceleration due to gravity (assumed to be 9.81 m/s²)

h is the depth of the ocean

Po is the atmospheric pressure at the surface (assumed to be 1.01 x 10^5 Pa)

Substituting the given values, we get:

P = (1.025 x 10^3 kg/m³) x (9.81 m/s²) x (1.0 x 10^3 m) + 1.01 x 10^5 Pa

P = 1.025 x 9.81 x 10^6 Pa + 1.01 x 10^5 Pa

P = 1.002 x 10^8 Pa.

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Explanation:

Given:

Diameter = 200 m

Radius, r = 200/2 = 100 m

Time taken, t = 30 seconds

Formula to be used:

Distance traveled, = circumference of circle = 2πr

Answer:

Putting all the values, we get

Distance traveled = 2πr

So, the distance traveled by Raju is 628.57 m.

Now, magnitude of the displacement,

At the end of 30 seconds, Raju will come to starting position or initial position, so displacement is zero.

The answer is that the time the diver was in the air is 1.13 seconds.

To determine the time the diver was in the air, we can use the kinematic equation:

Δy = viΔt + 1/2at²,

where Δy is the displacement, vi is the initial velocity, a is the acceleration due to gravity (g), and t is the time.The initial velocity, vi, is given as 5.5 m/s, and since the diver jumps upwards, the displacement, Δy, is equal to the height of the board, which is 3.0 m. The acceleration due to gravity, a, is -9.8 m/s² (negative because it acts downwards).Substituting the known values into the equation:3.0

m = (5.5 m/s)t + 1/2(-9.8 m/s²)t²

Simplifying, we get:

4.9t² + 5.5t - 3.0 = 0

We can solve for t using the quadratic formula:

t = (-5.5 ± √(5.5² - 4(4.9)(-3.0))) / (2(4.9))= (-5.5 ± 1.59) / 9.8= -0.47 s or 1.13 s

Since time cannot be negative, the time the diver was in the air is 1.13 seconds.

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Given,

The force exerted, F=30 N

The distance for which the force is applied, s=6.0 m

The mass of the object, m=10 kg

The initial velocity, u=0 m/s

From Newton's second law,

Thus the acceleration of the body is given by,

From the equation of the motion,

On substituting the known values, the final velocity of the object is

Thus the velocity of the object is 6 m/s

Answer:

D- moves with constant velocity

Explanation:

As we know that acceleration is tge rate of change in velocity.

Since velocity is constant, the change in velocity is zero (∆v=0)

\({ \tt{acceleration = \frac{ \triangle v}{t} }} \\ \\ { \tt{acceleration = \frac{0}{t} }} \\ \\ { \tt{acceleration = 0}}\)

Answer:

a = 0.62 [m/s²]

Explanation:

To solve this problem we must use the following equation of kinematics. But first, we must convert speeds from kilometers per hour to meters per second.

\(v_{f} =v_{o} -a*t\)

\(55[\frac{km}{hr}]*\frac{1hr}{3600s}*\frac{1000m}{1km} =15.27[\frac{m}{s} ]\\10[\frac{km}{hr} ]*\frac{1hr}{3600s}*\frac{1000m}{1km} = 2.77[\frac{m}{s} ]\)

where:

Vf = final velocity = 2,77 [m/s]

Vo = initial velocity = 15.27 [m/s]

t = time = 20 [s]

a = acceleration [m/s²]

Now replacing:

\(2.77=15.27-a*(20)\\20*a=12.49\\a = 0.62[m/s^{2}]\)

Answer:

It would be 10.00

Explanation:

Hope this helps its different for everyone what was it for u it was D for me

The weight of the Mars Rover Curiosity on Earth and on Mars is 8820 N and 3330 N respectively.

The weight of an object is given by the product of its mass and the gravitational field strength at its location.

On Earth:

On Mars:

Therefore, the weight of the Mars Rover Curiosity on Earth and on Mars are 8820 N and 3330 N respectively.

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Answer:

The more massive an object is, the greater the gravitational force. Since gravitational force is inversely proportional to the distance between two interacting objects, more separation distance will result in weaker gravitational forces.

Explanation:

hope this help:)

Answer:

C

Explanation:

Sublimation is when a solid turns into a gas.

A is not sublimation, it is condensation (gas to liquid = water vapor to dew)

B is evaporation (liquid to gas = liquid deodorant to gas)

D is melting (solid to liquid = ice cream to liquid ice cream)

E is condensation (gas to liquid = water vapor to water droplets)

C is the correct answer, because dry ice is a solid and carbon dioxide is a gas, so it is changing from solid to gas.

At point A the force is 571.6 N. At point B the force is 632.8 N. The minimum speed at point A is 8.32 m/s

The force of the car seat on the child can be determined using the equation F = m * a, where F is the force, m is the mass, and a is the acceleration. The acceleration can be found using the equation a = v^2/r, where v is the speed and r is the radius.
(a) At point a, the speed is 10.0 m/s and the radius is 7.00 m. Therefore, the acceleration is:
a = (10.0 m/s)^2 / (7.00 m) = 14.29 m/s^2
The force of the car seat on the child is:
F = (40.0 kg) * (14.29 m/s^2) = 571.6 N
(b) At point b, the speed is 10.5 m/s and the radius is 7.00 m. Therefore, the acceleration is:
a = (10.5 m/s)^2 / (7.00 m) = 15.82 m/s^2
The force of the car seat on the child is:
F = (40.0 kg) * (15.82 m/s^2) = 632.8 N
(c) The minimum speed required to keep the child in his seat at point a can be found by rearranging the equation for acceleration:
v = sqrt(a * r)
Since the force of the car seat on the child must be equal to or greater than the force of gravity on the child (F = m * g), the acceleration must be equal to or greater than the acceleration due to gravity (a = g = 9.81 m/s^2).
Therefore, the minimum speed is:
v = sqrt((9.81 m/s^2) * (7.00 m)) = 8.32 m/s

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a) The value of t u = 140/t`b.

b) The y position of the cannonball at the time t is  55.5 mc.

c) The initial speed of the projectile is 52.4 m/s.

Given that a cannonball is fired horizontally from the top of a cliff. The cannon is at height H = 55.5 m above ground level, and the ball is fired with initial horizontal speed u. The projectile lands at a distance D = 140 m from the cliff. Assume that the cannon is fired at time t = 0 and that the cannonball hits the ground at time t.Now,We have to find the value of t, y position of the cannonball at the time t and the initial speed of the projectile.

a. To find the value of t:Here, we have to use the formula of distance

i.e.,S = ut + (1/2)gt², Where S = 140 m, u = u and g = 9.8 m/s².Hence,140 = u×t ………..(1)We know that, time taken by the cannonball to hit the ground can be calculated as,`(2H)/g`

Since the height of the cannon from the ground is 55.5m, the total height of the cannonball from the ground is

(2H) = 2 × 55.5

= 111 m`2H/g

= 111/9.8`

= 11.32653 s

From equation (1),u×t = 140u = 140/t

Therefore, `u = 140/t`b.

b)To find the y position of the cannonball at the time t:

Here, we have to use the formula of height i.e.,y = u×t – (1/2)gt²,

Where, y = height of the cannonball at time t, u = 140/t, t = time taken by the cannonball to hit the ground and g = 9.8 m/s².

We have already calculated the time taken by the cannonball to hit the ground in the previous step.`

y = 140 - (1/2) × 9.8 × t²`

On substituting the value of t as `t = 11.32653`,

we get,y = 140 - (1/2) × 9.8 × (11.32653)²= 55.5 mc.

c)  To find the initial speed of the projectile:

To calculate the initial speed of the projectile, we need to use the formula of range of projectile

.i.e.,R = u²sin2θ/g

Where R = 140 m, g = 9.8 m/s², θ = 0° (horizontal)

u² = R × g/sin2θ

   = 140 × 9.8/sin0°  

   = 2744m²/s²u  

   = \(\sqrt(2744m^2/s^2)\)

   = 52.4 m/s

Hence, the initial speed of the projectile is 52.4 m/s.

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Answer:

13.94 rpm

Explanation:

Given that,

The diameter of the pulley, d = 14 foot

Radius, r = 7 foot

The linear velocity of the pulley, v = 14 mph = 20.53 ft/s

We need to find the angular velocity in rpm.

We know that, the relation between the linear velocity and the angular velocity is as follows :

\(v=r\omega\\\\\omega=\dfrac{v}{r}\\\\\omega=\dfrac{20.53}{14}\\\\\omega=1.46\ rad/s\)

or

\(\omega=13.94\ rpm\)

So, the angular velocity of the pulley is 13.94 rpm.

Answer:

P = 517.5 [kg*m/s]

Explanation:

We must remember that momentum is defined as the product of force by the time of force duration.

\(P=F*t\)

where:

P = momentum [kg*m/s]

F = force [N]

t = time [s]

\(P=345*1.5\\P=517.5[kg*m/s]\)