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3 years ago

11

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:

Fofino [41]3 years ago

7 0

Avg. acceleration = -5 m/s^2.

LuckyWell [14K]3 years ago

3 0

Answer:

Acceleration of the car is $-5\ m/s^2$

Explanation:

It is given that,

Initial speed of the car, u = 20 m/s

Finally, it brought to rest, v = 0

Time taken, t = 4 s

We need to find the average acceleration over this time interval. The acceleration of an object is given by following formula as :

$a=\dfrac{v-u}{t}$

$a=\dfrac{-(20\ m/s)}{4\ s}$

$a=-5\ m/s^2$

So, the car is decelerating at $5\ m/s^2$ . Hence, this is the required solution.

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If a skier travels at a constant velocity downhill, this means the forces on her are: ................. A. Zero B. Negligible C.

Vlad1618 [11]

Answer:

F = 0

Explanation:

The net force acting on an object is given by the product of mass and acceleration. We know that acceleration is equal to the rate of change of velocity.

Net force,

F = ma

$F=\dfrac{m(v-u)}{t}$

The skier is traveling at a constant velocity, it means there is no change in velocity i.e. acceleration is equal to 0. Hence, the force on her is 0.

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3 years ago

Although most magnetic reversing starters provide mechanical interlock protection, some circuits are provided with a secondary b

ioda

Answer:

To provide electrical interlocking

Explanation:

Electrical interlocking involves interconnecting the motor circuit in a manner that the second motor will not start until the first one begins, same goes for the third motor which would not run unless the second one runs and it continues in that sequence.

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3 years ago

a golfer tees off and hits a golf ball at a speed of 31 m/s and an angle of 35 degrees. how far did the ball travel before hitti

poizon [28]

Ans: R = Ball Travelled = 92.15 meters.

Explanation:
First we need to derive that formula for the "range" in order to know how far the ball traveled before hitting the ground.

Along x-axis, equation would be:
$x = x_o + v_o_xt + \frac{at^2}{2}$

Since there is no acceleration along x-direction; therefore,
$x = x_o + v_o_xt$

Since $v_o_x = v_ocos \alpha$ and $x_o$ =0; therefore above equation becomes,

$x = v_ocos \alpha t$ --- (A)

Now we need to find "t", and the time is not given. In order to do so, we shall use the y-direction motion equation. Before hitting the ground y ≈ 0 and a = -g; therefore,

=> $y = y_o + v_o_yt - \frac{gt^2}{2}$
=> $t = \frac{2v_o_y}{g}$

Since $v_o_y = sin \alpha$ ; therefore above equation becomes,
$t = \frac{2v_osin \alpha }{g}$

Put the value of t in equation (A):

(A) => $x = v_ocos \alpha \frac{2v_osin \alpha }{g}$

Where x = Range = R, and $2sin \alpha cos \alpha = sin(2 \alpha )$ ; therefore above equation becomes:

=> $R = (v_o)^2 *\frac{sin(2 \alpha )}{g}$

Now, as:
$v_o = 31 m/s$

and $\alpha = 35$ °
and g = 9.8 m/(s^2)

Hence,
$R = (31)^2 *\frac{sin(2 *35 )}{9.8}$

Ans: R = 92.15 meters.

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3 years ago

anthony is learning about electric circuits. he has started building a circuit shown below. which of the following items should

son4ous [18]

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3 years ago

An ore sample weighs 17.50 N in air. When the sample is suspended by a light cord and totally immersed in water, the tension in

valkas [14]

Answer:

Volume of the sample: approximately $\rm 0.6422 \; L = 6.422 \times 10^{-4} \; m^{3}$ .

Average density of the sample: approximately $\rm 2.77\; g \cdot cm^{3} = 2.778 \times 10^{3}\; kg \cdot m^{3}$ .

Assumption:

$\rm g = 9.81\; N \cdot kg^{-1}$ .
$\rho(\text{water}) = \rm 1.000\times 10^{3}\; kg \cdot m^{-3}$ .
Volume of the cord is negligible.

Explanation:

<h3>Total volume of the sample</h3>

The size of the buoyant force is equal to $\rm 17.50 - 11.20 = 6.30\; N$ .

That's also equal to the weight (weight, $m \cdot g$ ) of water that the object displaces. To find the mass of water displaced from its weight, divide weight with $g$ .

$\displaystyle m = \frac{m\cdot g}{g} = \rm \frac{6.30\; N}{9.81\; N \cdot kg^{-1}} \approx 0.642\; kg$ .

Assume that the density of water is $\rho(\text{water}) = \rm 1.000\times 10^{3}\; kg \cdot m^{-3}$ . To the volume of water displaced from its mass, divide mass with density $\rho(\text{water})$ .

$\displaystyle V(\text{water displaced}) = \frac{m}{\rho} = \rm \frac{0.642\; kg}{1.000\times 10^{3}\; kg \cdot m^{-3}} \approx 6.42201 \times 10^{-4}\; m^{3}$ .

Assume that the volume of the cord is negligible. Since the sample is fully-immersed in water, its volume should be the same as the volume of water it displaces.

$V(\text{sample}) = V(\text{water displaced}) \approx \rm 6.422\times 10^{-4}\; m^{3}$ .

<h3>Average Density of the sample</h3>

Average density is equal to mass over volume.

To find the mass of the sample from its weight, divide with $g$ .

$\displaystyle m = \frac{m \cdot g}{g} = \rm \frac{17.50\; N}{9.81\; N \cdot kg^{-1}} \approx 1.78389 \; kg$ .

The volume of the sample is found in the previous part.

Divide mass with volume to find the average density.

$\displaystyle \rho(\text{sample, average}) = \frac{m}{V} = \rm \frac{1.78389\; kg}{6.42201 \times 10^{-4}\; m^{3}} \approx 2.778\; kg \cdot m^{-3}$ .

3 0

3 years ago