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LuckyWell [14K]
2 years ago
14

A shopper does 157 J of work pushing a cart with 10.9 N force

Physics
1 answer:
Tanzania [10]2 years ago
4 0

The cart travelled a distance of 14.4 m

Explanation:

The work done by a force when pushing an object is given by:

W=Fd cos \theta

where:

F is the magnitude of the force

d is the displacement

\theta is the angle between the direction of the force and the displacement

In this problem we have:

W = 157 J is the work done on the cart

F = 10.9 N is the magnitude of the force

\theta=0^{\circ}, assuming the force is applied parallel to the motion of the cart

Therefore we can solve for d to find the distance travelled by the cart:

d=\frac{W}{F cos \theta}=\frac{157}{(10.9)(cos 0)}=14.4 m

Learn more about work:

brainly.com/question/6763771  

brainly.com/question/6443626  

#LearnwithBrainly

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"A light beam incident on a diffraction grating consists of waves with two different wavelengths. The separation of the two firs
Fantom [35]

Answer:

A light beam incident on a diffraction grating consists of waves with two different wavelengths. The separation of the two first order lines is great if

the dispersion is great

6 0
3 years ago
A toy rocket, launched from the ground, rises vertically with an acceleration of 28 m/s 2 for 9.7 s until its motor stops. Disre
vredina [299]

Answer:

5080.86m

Explanation:

We will divide the problem in parts 1 and 2, and write the equation of accelerated motion with those numbers, taking the upwards direction as positive. For the first part, we have:

y_1=y_{01}+v_{01}t+\frac{a_1t^2}{2}

v_1=v_{01}+a_1t

We must consider that it's launched from the ground (y_{01}=0m) and from rest (v_{01}=0m/s), with an upwards acceleration a_{1}=28m/s^2 that lasts a time t=9.7s.

We calculate then the height achieved in part 1:

y_1=(0m)+(0m/s)t+\frac{(28m/s^2)(9.7s)^2}{2}=1317.26m

And the velocity achieved in part 1:

v_1=(0m/s)+(28m/s^2)(9.7s)=271.6m/s

We do the same for part 2, but now we must consider that the initial height is the one achieved in part 1 (y_{02}=1317.26m) and its initial velocity is the one achieved in part 1 (v_{02}=271.6m/s), now in free fall, which means with a downwards acceleration a_{2}=-9,8m/s^2. For the data we have it's faster to use the formula v_f^2=v_0^2+2ad, where d will be the displacement, or difference between maximum height and starting height of part 2, and the final velocity at maximum height we know must be 0m/s, so we have:

v_{02}^2+2a_2(y_2-y_{02})=v_2^2=0m/s

Then, to get y_2, we do:

2a_2(y_2-y_{02})=-v_{02}^2

y_2-y_{02}=-\frac{v_{02}^2}{2a_2}

y_2=y_{02}-\frac{v_{02}^2}{2a_2}

And we substitute the values:

y_2=y_{02}-\frac{v_{02}^2}{2a_2}=(1317.26m)-\frac{(271.6m/s)^2}{2(-9.8m/s^2)}=5080.86m

3 0
3 years ago
How much power is required to light a lightbulb at 100V of voltage when the lightbulb has a resistance of 500 Ohms?
salantis [7]

Answer:

Power = 20 Watts

Explanation:

Given the following data;

Voltage = 100 V

Resistance = 500 Ohms

To find the power that is required to light a lightbulb;

Mathematically, power can be calculated using the formula;

Power = \frac {Voltage^{2}}{resistance}

Substituting into the formula, we have;

Power = \frac {100^{2}}{500}

Power = \frac {10000}{500}

Power = 20 Watts

3 0
3 years ago
Riders in a carnival ride stand with their backs against the wall of a circular room of diameter
Veseljchak [2.6K]

Answer:

option C

Explanation:

given,

diameter of circular room = 8 m

rotational velocity of the rider = 45 rev/min

                  = 45 \times \dfrac{2\pi}{60}

                  =4.712 rad/s

here in this case normal force is equal to centripetal force

N = m r ω²

N = m x 4 x 4.712²

N = 88.83m

frictional force = μ N

    = 88.83m x μ

now, for the body to not to slide

gravity force is equal to frictional force

m g = 88.83 m x μ

g = 88.83 x μ

9.8 = 88.83 x μ

 μ = 0.11

hence, the correct answer  is option C

6 0
3 years ago
A race car starting from rest accelerates uniformly at a rare of 4.90 meters per second^2. What is the cars speed after it has t
Vesnalui [34]

From the law of Galileo Galilei  :v²=v₀²+2ad we take the speed

v²=0+2*4.90*200=1960=>v=√1960=44.27 m/s




4 0
3 years ago
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