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2 years ago

13

A car is traveling at 20 meters/second and is brought to rest by applying brakes over a period of 4 seconds. What is its average

acceleration over this time interval?

2 answers:

frez [133]2 years ago

8 0

(u) = 20 m/s
(v) = 0 m/s
<span> (t) = 4 s
</span>
<span>0 = 20 + a(4)

</span><span>4 x a = -20
</span>
so, the answer is <span>-5 m/s^2. or -5 meter per second</span>

Feliz [49]2 years ago

8 0

The answer is -5 m/s^2. or -5 meter per second

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En el País Vasco los deportistas rurales levantan enormes piedras hasta su hombro. En un concurso , Jose levanta una piedra de 2

Mama L [17]

Answer:

Txomin lifted the stone with greater mass. (Txomin levantó la roca con mayor fuerza).

Explanation:

The sportsman that lifts the stone with a greater mass needs a higher force (El deportista que levanta la piedra con mayor masa necesita una mayor fuerza):

José

$F = (200\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)$

$F = 1961.4\,N$

Txomin

$F = (220\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)$

$F = 2157.54\,N$

Txomin lifted the stone with greater mass. (Txomin levantó la roca con mayor fuerza).

5 0

2 years ago

An object thrown in the air has a velocity after t seconds that can be described by v(t) = -9.8t + 24 (in meters/second) and a h

marin [14]

Answer and Explanation: Kinetic energy is related to movement: it is the energy an object possesses during the movement. it is calculated as:

$K=\frac{1}{2}mv^{2}$

For the object thrown in the air:

$K=\frac{1}{2}.2.[v(t)]^{2}$

$K=(-9.8t+24)^{2}$

$K=96.04t^{2}-470.4t+576$

Kinetic energy of the object as a function of time: $K=96.04t^{2}-470.4t+576$

Potential energy is the energy an object possesses due to its position in relation to other objects. It is calculated as:

$U=mgh$

For the object thrown in the air:

$U=9.8.2.h(t)$

$U=9.8.2.(-4.9t^{2}+24t+60)$

$U=-96.04t^{2}+470.4t+1176$

Potential energy as function of time: $U=-96.04t^{2}+470.4t+1176$

Total kinetic and potential energy, also known as mechanical energy is

TME = $96.04t^{2}-470.4t+576$ + ( $-96.04t^{2}+470.4t+1176$ )

TME = 1752

The expression shows that total energy of an object thrown in the air is constant and independent of time.

6 0

2 years ago

A 225 kg red bumper car is moving at 3.0 m/s. It hits a stationary 180 kg blue bumper car. The red and blue bumper cars combine

Alex Ar [27]

Given

m1(mass of red bumper): 225 Kg

m2 (mass of blue bumper): 180 Kg

m3(mass of green bumper):150 Kg

v1 (velocity of red bumper): 3.0 m/s

v2 (final velocity of the combined bumpers): ?

The law of conservation of momentum states that when two bodies collide with each other, the momentum of the two bodies before the collision is equal to the momentum after the collision. This can be mathemetaically represented as below:

Pa= Pb

Where Pa is the momentum before collision and Pb is the momentum after collision.

Now applying this law for the above problem we get

Momentum before collision= momentum after collision.

Momentum before collision = (m1+m2) x v1 =(225+180)x 3 = 1215 Kgm/s

Momentum after collision = (m1+m2+m3) x v2 =(225+180+150)x v2

=555v2

Now we know that Momentum before collision= momentum after collision.

Hence we get

1215 = 555 v2

v2 = 2.188 m/s

Hence the velocity of the combined bumper cars is 2.188 m/s

4 0

2 years ago

Read 2 more answers

1)After catching the ball, Sarah throws it back to Julie. However, Sarah throws it too hard so it is over Julie's head when it r

DENIUS [597]

Answer:

1)

$v_{oy}=11.29\ m/s$

2)

$y=7.39\ m$

Explanation:

<u>Projectile Motion</u>

When an object is launched near the Earth's surface forming an angle $\theta$ with the horizontal plane, it describes a well-known path called a parabola. The only force acting (neglecting the effects of the wind) is the gravity, which acts on the vertical axis.

The heigh of an object can be computed as

$\displaystyle y=y_o+V_{oy}t-\frac{gt^2}{2}$

Where $y_o$ is the initial height above the ground level, $v_{oy}$ is the vertical component of the initial velocity and t is the time

The y-component of the speed is

$v_y=v_{oy}-gt$

1) We'll find the vertical component of the initial speed since we have not enough data to compute the magnitude of $v_o$

The object will reach the maximum height when $v_y=0$ . It allows us to compute the time to reach that point

$v_{oy}-gt_m=0$

Solving for $t_m$

$\displaystyle t_m=\frac{v_{oy}}{g}$

Thus, the maximum heigh is

$\displaystyle y_m=y_o+\frac{v_{oy}^2}{2g}$

We know this value is 8 meters

$\displaystyle y_o+\frac{v_{oy}^2}{2g}=8$

Solving for $v_{oy}$

$\displaystyle v_{oy}=\sqrt{2g(8-y_o)}$

Replacing the known values

$\displaystyle v_{oy}=\sqrt{2(9.8)(8-1.5)}$

$\displaystyle v_{oy}=11.29\ m/s$

2) We know at t=1.505 sec the ball is above Julie's head, we can compute

$\displaystyle y=y_o+V_{oy}t-\frac{gt^2}{2}$

$\displaystyle y=1.5+(11.29)(1.505)-\frac{9.8(1.505)^2}{2}$

$\displaystyle y=1.5\ m+16,991\ m-11.098\ m$

$y=7.39\ m$

5 0

2 years ago

In an experiment, students roll several hoops down the same incline plane. Each hoop has the same mass but a different radius. E

insens350 [35]

Answer:

The graph should have velocity (v) on the y-axis and radius (r) on the x-axis. It will have a straight, horizontal line that goes across the graph.

Explanation:

$KE=\frac{1}{2} I(omega)^{2}$

Shown above is the formula for Kinetic Energy in rotational terms. I'm new to brain.ly so I couldn't insert the omega symbol, sorry about that. Omega can be replaced with $\frac{v^{2} }{r^2}$ . Moment of Inertia (I) can be replaced with $mr^2$ .

The equation becomes $KE=\frac{1}{2} mr^2(\frac{v^2}{r^2} )$ .

The r's cancel out, making the different radii negligible, causing a straight horizontal line.

5 0

2 years ago