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3 years ago

How does kinetic energy affect the stopping distance of a small vehicle compared to a large vehicle?

1 answer:

Studentka2010 [4]3 years ago

8 0

Kinetic energy is the energy applied or present in a moving object. According to Newton's second law of motion the magnitude of acceleration of an object is proportional to the magnitude of the net force but inversely proportional to its mass. So the Kinetic Energy of a moving car of small vehicle is greater than the large vehicle if both are applied with the same net force. The greater the Kinetic Energy the longer the stopping distance

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A Shaolin monk of mass 60 kg is able to do a ‘finger stand’: he supports his whole weight on his two index fingers, giving him a

LekaFEV [45]

Answer:

1500000 Pa

Explanation:

The formula for pressure is force per unit area.

P=F/A where F is force and A is area

Given that ;

F= mass * acceleration due to gravity

F= 60 * 9.81 = 588.6 = 589 N

A= area = 4cm² = 0.0004 m²

P= F/A = 589 / 0.0004

P= 1471500

P=1500000 Pa

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3 years ago

A boy throws a snowball straight up in the air with an initial speed of 4.50 ft/s from a position 4.00 ft above the ground. The

IgorC [24]

Answer:

a) 0.658 seconds

b) 0.96 inches

Explanation:

$v=u+at\\\Rightarrow 0=4.5-32.1\times t\\\Rightarrow \frac{-4.5}{-32.1}=t\\\Rightarrow t=0.14 \s$

Time taken by the ball to reach the highest point is 0.14 seconds

$s=ut+\frac{1}{2}at^2\\\Rightarrow s=4.5\times 0.14+\frac{1}{2}\times -32.1\times 0.14^2\\\Rightarrow s=0.315\ ft$

The highest point reached by the snowball above its release point is 0.315 ft

Total height the snowball will fall is 4+0.315 = 4.315 ft

$s=ut+\frac{1}{2}at^2\\\Rightarrow 4.315=0t+\frac{1}{2}\times 32.1\times t^2\\\Rightarrow t=\sqrt{\frac{4.315\times 2}{32.1}}\\\Rightarrow t=0.518\ s$

The snowball will reach the bank at 0.14+0.518 = 0.658 seconds after it has been thrown

$v=u+at\\\Rightarrow v=0+32.1\times 0.518\\\Rightarrow v=16.62\ ft/s$

$v^2-u^2=2as\\\Rightarrow s=\frac{v^2-u^2}{2a}\\\Rightarrow s=\frac{0^2-16.62^2}{2\times -100\times 3.28}\\\Rightarrow s=0.42\ ft$

The snowball goes 0.5-0.42 = 0.08 ft = 0.96 inches

8 0

3 years ago

Starting from rest, a disk rotates about its central axis with constant angular acceleration. In 5.00 s, it rotates 13.9 rad. Du

alisha [4.7K]

Answer:

(a) Angular acceleration is 1.112 rad/s².

(b) Average angular velocity is 2.78 rad/s .

Explanation:

The equation of motion in Rotational kinematics is:

θ = θ₀ + 0.5αt²

Here θ is angular displacement at time t, θ₀ is angular displacement at time t=0, t is time and α is constant angular acceleration.

(a) According to the problem, θ is 13.9 rad, θ₀ is zero as it is at rest and t is 5 s. Put these values in the above equation:

13.9 = 0 + 0.5α(5)²

α = 1.112 rad/s²

(b) The equation of average angular velocity is:

ω = Δθ/Δt

ω = $\frac{13.9}{5}$

ω = 2.78 rad/s

3 0

3 years ago

Which statement is true regarding a distance vs. time graph? (1 point)

Nadusha1986 [10]

The statement that is true regarding a distance vs. time graph is option A: The graph should show distance on the vertical axis.

<h3>Where is the plot of distance?</h3>

How far an object has come in a certain amount of time is displayed on a distance-time graph. Time is represented on the X-axis and Distance is plotted on the Y-axis (left) (bottom).

On a distance-time graph, an object's motion is indicated by a sloping line. The slope or gradient of the line in a distance-time graph is equal to the object's speed. The object is travelling more quickly the steeper the line is (and the bigger the gradient).

Note that the distance-time graph shows the relationship between distance and time by plotting distance on the y-axis and time on the x-axis.

Learn more about distance vs. time graph from

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