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4 years ago

Why is weight considered a force

1 answer:

4 0

Because if weight is being applied to an object, the object will have some sort of effect to counter react due to the weights primary objective.

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A mountain climber increases their height from 200 meters to 400 meters. What affect will this have on their potential energy?

Yanka [14]

Answer:

At 400 m the potential energy of the mountain climber doubled the initial value.

Explanation:

Given;

initial height of the mountain climber = 200 m

final height of the mountain climber, = 400 m

The potential energy of the mountain climber is calculated as;

Potential energy, P.E = mgh

At 200 m, P.E₁ = mg x 200 = 200mg

At 400 m, P.E₂ = mg x 400 = 400mg

Then, at 400 m, P.E₂ = 2 x 200mg = 2 x P.E₁

Therefore, at 400 m the potential energy of the mountain climber doubled the initial value.

4 0

3 years ago

During a race, a runner runs at a speed of 6 m/s. Four seconds later, she is running at a speed of 10 m/s. what is the runners a

Georgia [21]

<span>a = (v2 - v1)/t = (10 - 6)/2 = 2 m/sec/sec (average acceleration)</span>

8 0

4 years ago

A non-conducting sphere of radius R = 3.0 cm carries a charge Q = 2.0 mC distributed uniformly throughout its volume. At what di

OleMash [197]

To solve this problem we will use the concept of electric field, with which we will make the proportional comparison as we move away from the center. So we have the maximum electric field is given as,

$E_{max} = \frac{kQ}{R^2}$

Where,

Q = Charge

R = Radius

Electric field inside the sphere is given as,

$\frac{kQ}{R^2} = \frac{1}{2}\frac{kQ}{R^2}$

$R'=\sqrt{2}R$

$R' = 3\sqrt{2}$

$R' = 4.2cm$

Electric field outside the sphere is given as,

$\frac{kQ}{2R^2} = \frac{1}{2}\frac{kQ}{R^3}r$

$\frac{1}{2} = \frac{r'}{R}$

$\Rightarrow \frac{R}{2} = \frac{3}{2} = 1.5cm$

Therefore the possible values are 3.5cm and 9.9cm: The correct answer is D.

5 0

3 years ago

A car moving at 10.0 m/s encounters a bump that has a circular cross-section with a radius of 30.0 m. What is the normal force e

Andrew [12]

Answer:

force at the top will be 388 N which is nearly equal to 389

option (b) is correct

Explanation:

We have given velocity of the moving car v = 10 m/sec

Radius of the circular section r = 30 m

Mass of the passenger m = 60 kg

Acceleration due to gravity $g=9.8m/sec^2$

At the top normal force is given by $noraml\ force=mg-\frac{mv^2}{r}$

So force at top will be $F=60\times 9.8-\frac{60\times 10^2}{30}=388N$

So force at the top will be 388 N which is nearly equal to 389

So option (b) is correct

4 0

3 years ago

Read 2 more answers

A student throws a set of keys vertically upward to his fraternity brother, who is in a window 4.00 m above. The brother’s outst

MakcuM [25]

Answer:10.02 m/s

4.68 m/s

Explanation:

Given

height of building=4 m

time taken=1.5 s

(a)Let u be the initial velocity

using equation of motion

$s=ut+\frac{gt^2}{2}$

$4=u\times 1.5-\frac{9.81\times 1.5^2}{2}$

$8=u\times 3-9.81\times 2.25$

u=10.02 m/s

(b)Velocity of keys just before keys were caught

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

$v^2=10.02^2-2(-9.81)\cdot 4$

$v^2=21.92$

$v=\sqrt{21.92}=4.68 m/s$

3 0

3 years ago