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2 years ago

Jake and Reuben are practicing for their weightlifting meet at school.

Physics

1 answer:

hammer [34]2 years ago

5 0

Answer:

no Jake arm is high so the weight of an object become high due to potential energy.

so, option A is correct ) Jake did more work

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Isla’s change in velocity is 30 m/s, and Hazel has the same change in velocity. Which best explains why they would have differen

irina [24]

Answer:

Explanation:

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3 0

3 years ago

Read 2 more answers

In our first example we will consider a very simple application of Newton’s second law. A worker with spikes on his shoes pulls

sweet-ann [11.9K]

Answer:

Acceleration= $0.5m/s^2$

Speed=0.67 m/s

Explanation:

We are given that

Horizontal force=F=20 N

Mass of box=m=40 kg

We know that

Acceleration= $a=\frac{F}{m}$

Using the formula

Acceleration of box= $\frac{20}{40}=0.5m/s^2$

The acceleration of the box= $0.5m/s^2$

Initial velocity=u=0

Force=F=30 N

Distance=s=0.3 m

$a=\frac{30}{40}=\frac{3}{4} ms^{-2}$

$v^2-u^2=2as$

Substitute the values

$v^2-0=2\times \frac{3}{4}\times 0.3=0.45$

$v^2=0.45$

$v=\sqrt{0.45}=0.67m/s$

Hence, the speed of the box after it has been pulled a distance of 0.3 m=0.67 m/s

4 0

3 years ago

You want to build a search light. The search light is to have an outgoing parallel beam. You have a small, very bright light (a

andrew11 [14]

Answer:

1 m

Explanation:

Given

The mirror is a concave mirror.

Radius of curvature = 2 m

Now, for a parallel beam after reflection occurs, the object should be placed at the focus of the mirror.

We know, focal length = radius of curvature / 2

= 2 / 2

= 1

Thus the focal length is 1 m.

Hence the object should be placed at 1 m from the concave mirror for parallel beam reflection.

3 0

3 years ago

In January 2004, NASA landed exploration vehicles on Mars. Part of the descent consisted of the following stages:

fenix001 [56]

Acceleration is given by:

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

where

v is the final velocity

u is the initial velocity

t is the time interval

Let's apply the formula to the different parts of the problem:

A) $-20.5 m/s^2$

Let's convert the quantities into SI units first:

$u = 19300 km/h \cdot \frac{1000 m/km}{3600 s/h} = 5361.1 m/s$

$v=1600 km/h \cdot \frac{1000 m/km}{3600 s/h} =444.4 m/s$

t = 4.0 min = 240 s

So the acceleration is

$a=\frac{444.4 m/s-5361.1 m/s}{240 s}=-20.5 m/s^2$

B) $-3.8 m/s^2$

As before, let's convert the quantities into SI units first:

$u = 444.4 m/s$

$v=321 km/h \cdot \frac{1000 m/km}{3600 s/h} =89.2 m/s$

t = 94 s

So the acceleration is

$a=\frac{89.2 m/s - 444.4 m/s}{94 s}=-3.8 m/s^2$

C) $-53.0 m/s^2$

For this part we have to use a different formula:

$v^2 - u^2 = 2ad$

where we have

v = 0 is the final velocity

u = 89.2 m/s is the initial velocity

a is the acceleration

d = 75 m is the distance covered

Solving for a, we find

$a=\frac{v^2-u^2}{2d}=\frac{0^2-(89.2 m/s)^2}{2(75 m)}=-53.0 m/s^2$

3 0

3 years ago

What is meant by the phrase "a consistent method of measurement"?

julia-pushkina [17]

A measurement that will always give the same answer.

5 0

3 years ago