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nataly862011 [7]
3 years ago
12

PLEASE HELP!! WILL GIVE POINTS

Physics
1 answer:
AURORKA [14]3 years ago
3 0

Answer:

49 kg is the mass of the couch.

Explanation:

GPE = mgh

9800 = m * 10 * 20

9800 = 200m

m = (9800/20) = 49 m

Thenks and mark me brainliest :))

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1) Manipulate is to measure as
ValentinkaMS [17]

Answer: The variable that you manipulate is called the independent variable. The variable that you measure is called the dependent variable.

3 0
3 years ago
Determine the amount of torque
malfutka [58]

Answer:

250Nm

Explanation:

Given parameters:

Length of the long pry bar  = 1m

Force acting on it  = 250N

Angle  = 90°

Unknown:

Amount of torque applied  = ?

Solution:

Torque is the turning force on a body that causes the rotation of the body.

The formula is given as:

 Torque  = Force x r Sin Ф  

r is the distance

 So;

   Torque  = 250 x 1 x sin 90  = 250Nm

7 0
3 years ago
A roller coaster speeds up with constant acceleration for 2.3 s until it reaches a velocity of 35 m/s. During this time, the rol
Arada [10]

The initial velocity is 0.65 m/s

Explanation:

The motion of the roller coaster is a uniformly accelerated motion, so we can solve the problem by using the following suvat equation:

s=(\frac{u+v}{2})t

where

s is the displacement

u is the initial velocity

v is the final velocity

t is the time

For the roller coaster in this problem:

v = 35 m/s

t = 2.3 s

s = 41 m

Solving the equation for u, we find the initial velocity:

u=\frac{2s}{t}-v=\frac{2(41)}{2.3}-35=0.65 m/s

Learn more about accelerated motion:

brainly.com/question/9527152

brainly.com/question/11181826

brainly.com/question/2506873

brainly.com/question/2562700

#LearnwithBrainly

5 0
3 years ago
a new planet is discovered that has twice the earth’s mass and twice the earth’s radius. on the surface of this new planet, a pe
Irina18 [472]

Answer: 250n

Explanation:

The formula for gravitational force is: F = (gMm)/r^2

There are two factors at play here:

1) The mass of the planet 'M'

2) The radius 'r'

We can ignore the small M and the g, they are constants that do not alter the outcome of this question.

You can see that both M and r are double that of earth. So lets say earth has M=1 and r=1. Then, new planet would have M=2 and r=2. Let's sub these two sets into the equation:

Earth. F =  M/r^2 = 1/1

New planet. F = M/r^2 = 2/4 = 1/2

So you can see that the force on the new planet is half of that felt on Earth.

The question tells us that the force on earth is 500n for this person, so then on the new planet it would be half! So, 250n!

8 0
2 years ago
A 1,508 kg car rolling on a horizontal surface has a speed of 20.8 km/hr when it strikes a horizontal coiled spring and is broug
natali 33 [55]

Answer:

Approximately 1.79 \times 10^{5}\; {\rm N \cdot m^{-1}}, assuming friction between the vehicle and the ground is negligible.

Explanation:

Let m denote the mass of the vehicle. Let v denote the initial velocity of the vehicle. Let k denote the spring constant (needs to be found.) Let x denote the maximum displacement of the spring.

Convert velocity of the vehicle to standard units (meters per second):

\begin{aligned}v &= 20.8\; {\rm km \cdot h^{-1}} \times \frac{1000\; {\rm m}}{1\; {\rm km}} \times \frac{1\; {\rm h}}{3600\; {\rm s}} \\ &\approx 1.908\; {\rm m \cdot s^{-1}}\end{aligned}.

Initial kinetic energy ({\rm KE}) of the vehicle:

\begin{aligned}\frac{1}{2}\, m \, v^{2}\end{aligned}.

When the vehicle is brought to a rest, the elastic potential energy (\text{EPE}) stored in the spring would be:

\displaystyle \frac{1}{2}\, k\, x^{2}.

By the conservation of energy, if the friction between the vehicle and the ground is negligible, the initial \text{KE} of the vehicle should be equal to the {\rm EPE} of the vehicle. In other words:

\begin{aligned}\frac{1}{2}\, m \, v^{2} &= \frac{1}{2}\, k\, x^{2}\end{aligned}.

Rearrange this equation to find an expression for k, the spring constant:

\begin{aligned}k &= \frac{m\, v }{x^{2}}\end{aligned}.

Substitute in the given values m = 1508\; {\rm kg}, v \approx 1.908\; {\rm m\cdot s^{-1}}, and x = 6.87\; {\rm m}:

\begin{aligned}k &= \frac{m\, v }{x^{2}} \\ &\approx \frac{1508\; {\rm kg} \times 1.908\; {\rm m\cdot s^{-1}}}{(6.87\; {\rm m})^{2}} \\ &\approx 1.79 \times 10^{5}\; {\rm kg \cdot m \cdot s^{-2} \cdot m^{-3}}\\ &\approx 1.79 \times 10^{5}\; {\rm N \cdot m^{-1}}\end{aligned}

8 0
2 years ago
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