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Agata [3.3K]
3 years ago
14

If the wagon travels 18.75 m, what is the work done on the wagon

Physics
1 answer:
Schach [20]3 years ago
4 0

If the wagon travels 18.75 m, then the work done on the wagon is

(18.75 m) x (the steady force applied to the wagon all the way, in Newtons) .

The unit is Joules .


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Kepler’s third law states that for all objects orbiting a given body, the cube of the semimajor axis (A) is proportional to the
GarryVolchara [31]

Explanation:

Kepler’s third law states that for all objects orbiting a given body, the cube of the semimajor axis (A) is proportional to the square of the orbital period (P).

For each of our planets orbiting the Sun, the relationship between the orbital period and semimajor axis can be represented by the equation as:

P^2=kA^3

k is constant of proportionality

It is required to solve the above equation for k

k=\dfrac{P^2}{A^3}

8 0
3 years ago
Starting from rest, a dragster travels a straight 1/4 mi racetrack in 7.10 s with constant acceleration. What is its velocity wh
Gennadij [26K]

268.6567 mph  is its velocity when it crosses the finish line

d=(v1+v2 /2) x t

.25=(0+v2 /2) x 6.7/3600 hours

900=v2/2 x 6.7

v2=268.6567 mph as the speed with which the dragster crosses the finish

<h3>When acceleration is not zero, can speed remain constant?</h3>

The answer is that an accelerated motion can have a constant speed. Consider a particle travelling uniformly around a circle; it experiences acceleration since the motion's direction is changing, but it maintains a constant speed along the tangential axis throughout the motion.

Acceleration is the frequency of a change in velocity. Acceleration is a vector with magnitude and direction, much as velocity. For instance, if a car is moving in a straight path and speeding up, it is said to have forward (positive) acceleration, and if it is slowing down, it is said to have backward (negative) acceleration.

Learn more about velocity refer

brainly.com/question/24681896

#SPJ9

5 0
1 year ago
How fast is a car going if it accelerates at 10m/s/s for 4 seconds
lara31 [8.8K]

Answer:

40 ms-1

Explanation:

v=u+at

v=0+10×4

v=40

5 0
3 years ago
There are two parallel conductive plates separated by a distance d and zero potential. Calculate the potential and electric fiel
taurus [48]

Answer:

The total electric potential at mid way due to 'q' is \frac{q}{4\pi\epsilon_{o}d}

The net Electric field at midway due to 'q' is 0.

Solution:

According to the question, the separation between two parallel plates, plate A and plate B (say)  = d

The electric potential at a distance d due to 'Q' is:

V = \frac{1}{4\pi\epsilon_{o}}.\frac{Q}{d}

Now, for the Electric potential for the two plates A and B at midway between the plates due to 'q':

For plate A,

V_{A} = \frac{1}{4\pi\epsilon_{o}}.\frac{q}{\frac{d}{2}}

Similar is the case with plate B:

V_{B} = \frac{1}{4\pi\epsilon_{o}}.\frac{q}{\frac{d}{2}}

Since the electric potential is a scalar quantity, the net or total potential is given as the sum of the potential for the two plates:

V_{total} = V_{A} + V_{B} = \frac{1}{4\pi\epsilon_{o}}.q(\frac{1}{\frac{d}{2}} + \frac{1}{\frac{d}{2}}

V_{total} = \frac{q}{4\pi\epsilon_{o}d}

Now,

The Electric field due to charge Q at a distance is given by:

\vec{E} = \frac{1}{4\pi\epsilon_{o}}.\frac{Q}{d^{2}}

Now, if the charge q is mid way between the field, then distance is \frac{d}{2}.

Electric Field at plate A, \vec{E_{A}} at midway due to charge q:

\vec{E_{A}} = \frac{1}{4\pi\epsilon_{o}}.\frac{q}{(\frac{d}{2})^{2}}

Similarly, for plate B:

\vec{E_{B}} = \frac{1}{4\pi\epsilon_{o}}.\frac{q}{(\frac{d}{2})^{2}}

Both the fields for plate A and B are due to charge 'q' and as such will be equal in magnitude with direction of fields opposite to each other and hence cancels out making net Electric field zero.

3 0
3 years ago
VELOCITY, SPEED, DISTANCE ETC..
vaieri [72.5K]

Answer:

Explanation:

Speed = distance / time

Velocity  = displacement / time

So ,

Speed = 50 km / 0.5 hr = 100 km/h

Velocity  = 40 km / 0.5hr = 80 km/h

4 0
2 years ago
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