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A shopper pushes a grocery cart 20.0 m at constant speed on level ground, against a 35.0 N frictional force. He pushes in a dire

svp [43]

Answer:

(A). The work done on the cart by the friction is -700 J.

(B). The work done on the cart by the gravitational force is 0 J.

(C). The work done on the cart by the shopper is 700 J.

(D). The force the shopper exerts 38.61 N.

(E). The total work done on the cart is zero.

Explanation:

Given that,

Distance = 20.0 m

Frictional force = 35.0 N

Angle = 25.0°

(A). We need to calculate the work done on the cart by the friction

Using formula of work done

$W_{fr} = -F\cdot d$

Where, F = force

d = distance

Put the value into the formula

$W_{fr}=-35.0\times20$

$W_{fr}=−700\ J$

(B). The work done by the gravity is perpendicular to the direction of the motion

We need to calculate the work done on the cart by the gravitational force

Using formula of work done

$W=fd\cos\theta$

Put the value into the formula

$W=35.0\times20\cos90$

$W=0\ J$

(C). We need to calculate the work done on the cart by the shopper

Using formula of work done

$W_{sh}=W_{net}-W_{fr}$

Put the value into the formula

$W_{sh}=0-(-700)$

$W_{sh}=700\ J$

(D). We need to calculate the force the shopper exerts

Using formula of force

$F_{sh}=\dfrac{W_{fr}}{d\cos\theta}$

Put the value into the formula

$F_{sh}=\dfrac{700}{20\cos25}$

$F_{sh}=38.61\ N$

(E). We need to calculate the total work done on the cart

Using formula of work done

$W_{cart}=W_{fr}+W_{sh}$

Put the value into the formula

$W_{cart}=700-(-700)$

$W_{cart}=0\ J$

Hence, (A). The work done on the cart by the friction is -700 J.

(B). The work done on the cart by the gravitational force is 0 J.

(C). The work done on the cart by the shopper is 700 J.

(D). The force the shopper exerts 38.61 N.

(E). The total work done on the cart is zero.

6 0

3 years ago

An object is released from rest and falls a distance h during the first second of time. How far will it fall during the next sec

Viefleur [7K]

Answer:

E. 3h

Explanation:

We know that

u = 0 m/s.

velocity after t = 1s

v = u+gt = 0+9.81 x 1s= 9.81 m/s

distance covered in 1st sec

= =>> ut+0.5 x g x t²

=>>0 + 0.5x 9.81 x 1 = 4.90m

Let 4.90 be h

distance travelled in 2nd second will now be used

So velocity after t = 1s

=>>1 x t+ 0.5 x g x t²

=>9.81x 1 + 0.5 x 9.81 x 1 = 3 x 4.90

So since h= 4.90

Then the ans is 3x h = 3h

3 0

3 years ago

A spaceship accelerates uniformly for 1220km how much time is needed for the spaceship to increase its speed from 11.1km/s to 11

snow_lady [41]

The time taken for the spaceship to increase its speed from 11.1 km/s to 11.7 km/s is 107 s

<h3>Data obtained from the question</h3>

The following data were obtained from the question given above:

Initial velocity (u) = 11.1 Km/s
Final velocity (v) = 11.7 Km/s
Distance (s) = 1220 Km
Time (t) =?

<h3>How to determine the time</h3>

The time taken for the spaceship to increase its speed from 11.1 km/s to 11.7 km/s can be obtained as illustrated below:

s = (u + v)t / 2

Cross multiply

(u + v)t = 2s

Divide both sides by (u + v)

t = 2s / (u + v)t

t = (2 × 1220) / (11.1 + 11.7)

t = 2440 / 22.8

t = 107 s

Thus, the time taken for the spaceship to change its speed is 107 s

Learn more about speed:

brainly.com/question/680492

Learn more about velocity:

brainly.com/question/3411682

#SPJ1

6 0

2 years ago

Suppose that you are swimming in a river while a friend watches from the shore. In calm water, you swim at a speed of 1.25 m/s .

aliya0001 [1]

Answer: The observing friend will the swimmer moving at a speed of 0.25 m/s.

Explanation:

Let <em>S</em> be the speed of the swimmer, given as 1.25 m/s

Let $S_{0}$ be the speed of the river's current given as 1.00 m/s.

Note that this speed is the magnitude of the velocity which is a vector quantity.

The direction of the swimmer is upstream.

Hence the resultant velocity is given as, $S_{R}$ = S — S 0 $S_{0}$

$S_{R}$ = 1.25 — 1

$S_{R}$ = 0.25 m/s.

Therefore, the observing friend will see the swimmer moving at a speed of 0.25 m/s due to resistance produced by the current of the river.

6 0

3 years ago

A vegatable garden is 12 meters long by 7 meters wide.It is home to 168 mice.What is the population density of the mice?

chubhunter [2.5K]

Density= population/area
area= 12 X 7 = 84 m^2
density= 168/84= 2 mice per m^2

5 0

4 years ago