(25 Brainly Points.)

What force must be used to do 224 Joules of work on an object over a distance of 32 meters?

cestrela7 [59]

7.625 Newtons

work = force× distance
Newtons is an accepted value for force

so take the total 224 joules and decide by distance 32 meters to find force in Newtons

4 0

3 years ago

An artillery shell is fired with an initial velocity of 300 m/s at 52.0° above the horizontal. To clear an avalanche, it explode

sasho [114]

The x- and y-coordinates are 9142.57 m and -304.425 m

<u>Explanation:</u>

As the motion of the shell is in a plane (two dimensional space) and the acceleration is that due to gravity which is vertically downward, we resolve initial velocity of the shell $v_{0}$ in horizontal and vertical directions. If the initial velocity of the shell is making angle with the horizontal, the horizontal component of initial velocity will be

$v_{x}=v_{0} \times \cos \theta$

As the acceleration of the shell is vertical having no horizontal component, the shell may be considered to move horizontally with constant velocity of $v_{x}$ and hence the horizontal distance covered (or the x coordinate of the shell with point of projection as origin) is given by

$v_{x}=v_{o} \times \cos \theta=300 \times \cos \left(52^{\circ}\right)=184.69 \mathrm{m} / \mathrm{s}$

$v_{y}=v_{o} \times \sin \theta==300 \times \sin \left(52^{\circ}\right)=236.4 \mathrm{m} / \mathrm{s}$

For motion with constant acceleration, we know

$s=s_{0}+v_{0} t+\left(\frac{(1)}{2}\right) a t^{2}$

Along the horizontal, x-axis, we might write this as

$x=x_{0}+v_{x 0} t+\left(\frac{1}{2}\right) a_{x} t^{2}$

Measuring distances relative to the firing point means

$x_{0}=0$

we know that,

$a_{x}=0$

or,

$v_{x}=v_{x 0}=\text { constant }$

By applying the values, we get,

$x=0+(184.69 \times 49.5)+\left(\left(\frac{1}{2}\right) \times 0 \times(49.5)^{2}\right)=9142.57 \mathrm{m}$

The acceleration of gravity is vertically downward and is $g=-9.8 \mathrm{m} / \mathrm{s}^{2}$ , hence the vertical distance covered (or y coordinate of the shell) is given by the second equation of motion

$y=y_{0}+v_{y 0} t+\left(\frac{1}{2}\right) a_{y} t^{2}$

we know, $y_{0}=0$ and $a_{y}=-9.8 \mathrm{m} / \mathrm{s}^{2}$ , so,

$y=0+(236.4 \times 49.5)+\left(\left(\frac{1}{2}\right) \times(-9.8) \times(49.5)^{2}\right)$

y = 11701.8 - 4.9(2450.25)= 11701.8 - 12006.225 = - 304.425 m

7 0

3 years ago

Joey is riding in an elevator which is accelerating upwards at 2.0 m/s2. The elevator weighs 300.0 kg, and Joey weighs 60.0 kg.

vodka [1.7K]

4200 N is the tension in the cable that pulls the elevator upwards.

The correct option is A.

<h3>What does tension ?</h3>

Tension is the force that is sent through a rope, thread, or wire whenever two opposing forces pull on it. Along the whole length of the wire, the tensile stress pulls equally on all objects at the ends. Every physical object that comes into contact with that other one exerts force on it.

<h3>Briefing:</h3>

We employ the following formula to determine the cable's tension.

Formula:

T = mg+ma............ Equation 1

Where:

T is the cable's tension.

M = Mass of the elevator and the Joey

Accelerating with a

g = Gravitational acceleration

Considering the query,

Given:

m = (300+60) = 360 kg

a = 2 m/s²

g = 9.8 m/s²

Substitute these values into equation 2

T = (360×9.8)+(360×2)

T = 3528+720

T = 4248 N

T ≈ 4200 to the nearest hundred.

To know more about Tension visit:

brainly.com/question/14177858

#SPJ1

7 0

1 year ago

A weightlifter presses a 200 N weight 0.5 m over his head in 2 s. What is the power of the weightlifter

icang [17]

Answer:

50 watts

Explanation:

Applying,

Power (P) = Workdone (W)/Time(t)

But,

Work done (W) = Force (F)×distance(d)

Therefore,

P = Fd/t..................... Equation 1

Where P = power of the weightlifter, F = Force applied, d = distance, t = time.

From the question,

Given: F = 200 N, d = 0.5 m, t = 2 s

Substitute these values into equation 1

P = (200×0.5)/2

P = 100/2

P = 50 watts

7 0

2 years ago

A player holds two baseballs a height h above the ground. He throws one ball vertically upward at speed v0 and the other vertica

Degger [83]

Answer:

a) v = √(v₀² + 2g h), b) Δt = 2 v₀ / g

Explanation:

For this exercise we will use the mathematical expressions, where the directional towards at is considered positive.

The velocity of each ball is

ball 1. thrown upwards vo is positive

v² = v₀² - 2 g (y-y₀)

in this case the height y is zero and the height i = h

v = √(v₀² + 2g h)

ball 2 thrown down, in this case vo is negative

v = √(v₀² + 2g h)

The times to get to the ground

ball 1

v = v₀ - g t₁

t₁ = $\frac{v_{o} - v }{ g}$

ball 2

v = -v₀ - g t₂

t₂ = - \frac{v_{o} + v }{ g}

From the previous part, we saw that the speeds of the two balls are the same when reaching the ground, so the time difference is

Δt = t₂ -t₁

Δt = $\frac{1}{g} \ [(v_{o} - v) - ( - v_{o} - v) ]$

Δt = 2 v₀ / g

6 0

3 years ago