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3 years ago

What is the value of work done on an object when a 50–newton force moves it 15 meters in the same direction as the force?

Physics

1 answer:

laila [671]3 years ago

4 0

Answer:

Work done, W = 750 joules

Explanation:

It is given that,

Force acting on the object, F = 50 N

It moves to a distance of, d = 15 meters

We need to find the work done on an object. We know that the product of force and distance covered is called the work done. As the force and the displacement are in same direction. So,

$W=F\times d$

$W=50\ N\times 15\ m$

W = 750 joules

So, the work done on an object is 750 joules. Hence, this is the required solution.

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Arace car accelerates uniformly at 11.3 m/s2. If the race car starts from rest how fast will it

Ganezh [65]

Answer:

75.71 m/s

Explanation:

From equation of motion, acceleration is given by

$a=\frac {v-u}{t}$ where v is the final velocity, u is the initial velocity and t is time taken.

Making v the subject of the above formula

v=at+u

Substituting 6.7 s for time, t and 11.3 for a and taking u as zero since it starts from rest

v=11.3*6.7=75.71 m/s

7 0

3 years ago

A baby is dropped from a 30 m cliff on Earth. Acceleration due to gravity on Earth is about 9.8 m/s^2. Calculate the following q

Katarina [22]

Answer:Dropping is not a good predictor of when labor will begin. In first-time mothers, dropping usually occurs 2 to 4 weeks before delivery, but it can happen earlier. In women who have already had children, the baby may not drop until labor begins.

Explanation:

4 0

3 years ago

As mentioned in the text, the tangent line to a smooth curve r(t) = ƒ(t)i + g(t)j + h(t)k at t = t0 is the line that passes thro

LiRa [457]

Answer:

$x = t$

$y = \frac{1}{3}t$

$z =t$

Explanation:

Given

$r(t) = f(t)i + g(t)j + h(t)k$ at $t = 0$

Point: $(f(t0), g(t0), h(t0))$

$r(t) = ln\ t_i + \frac{t-1}{t+2}j + t\ ln\ tk$ , $t0 = 1$ -- Missing Information

Required

Determine the parametric equations

$r(t) = ln\ ti + \frac{t-1}{t+2}j + t\ ln\ tk$

Differentiate with respect to t

$r'(t) = \frac{1}{t}i +\frac{3}{(t+2)^2}j + (ln\ t + 1)k$

Let t = 1 (i.e $t0 = 1$ )

$r'(1) = \frac{1}{1}i +\frac{3}{(1+2)^2}j + (ln\ 1 + 1)k$

$r'(1) = i +\frac{3}{3^2}j + (0 + 1)k$

$r'(1) = i +\frac{3}{9}j + (1)k$

$r'(1) = i +\frac{1}{3}j + (1)k$

$r'(1) = i +\frac{1}{3}j + k$

To solve for x, y and z, we make use of:

$r(t) = f(t)i + g(t)j + h(t)k$

This implies that:

$r'(1)t = xi + yj + zk$

So, we have:

$xi + yj + zk = (i +\frac{1}{3}j + k)t$

$xi + yj + zk = it +\frac{1}{3}jt + kt$

By comparison:

$xi = it$

Divide by i

$x = t$

$yj = \frac{1}{3}jt$

Divide by j

$y = \frac{1}{3}t$

$zk = kt$

Divide by k

$z = t$

Hence, the parametric equations are:

$x = t$

$y = \frac{1}{3}t$

$z =t$

3 0

3 years ago

Between which two planets is the gravitational attraction the weakest?

raketka [301]

planet B and D have the weakest gravitational attraction.

5 0

3 years ago

Read 2 more answers

What two physical processes balance each other to create the condition known as gravitational equilibrium in stars?

dusya [7]

Answer:

gravitational force and outward pressure

Explanation:

The stars are very massive stellar objects, so they have a strong gravitational force, which drives the star to contract itself, but also in stars there are nuclear reactions such as fusion of hydrogen and other elements, that releases energy and creates a pressure from the center to the star exterior, an outward pressure that goes against the gravitational force. So when a star is stable these two forces exist in equilibrium or in balance, in which the star does not collapse by gravity or disintegrate by its outward pressure.

3 0

3 years ago