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3 years ago

the weight of an object on the earths surface is 300N. when it is lifted to 3 times the height, its weight will become

Physics

1 answer:

emmainna [20.7K]3 years ago

6 0

Answer:

The weight of the object when it is lifted to 3 times the height is 33. $\overline 3$ N

Explanation:

The given parameters are;

The weight of the object on Earth, W = 300 N

The initial position of the object = On the surface of the Earth

Therefore;

The distance with which the weight is measured = The radius of the Earth, R

By Newton's Law of Gravitation, we have;

$W = G \times \dfrac{M \times m}{R^2}$

Where;

W = 200 N

G = The universal gravitational constant

M = The mass of the Earth

m = The mass of the object

When the height of the object = 3 × R, the weight of the object, W₂, is given as follows;

$W_2 = G \times \dfrac{M \times m}{(3 \times R)^2} = \dfrac{1}{9} \times G \times \dfrac{M \times m}{ R^2} = \dfrac{1}{9} \times W = \dfrac{1}{9} \times 300 N = 33.\overline 3 \ N$

Therefore, the weight of the object at 3 times the height, W₂ = 33. $\overline 3$ N

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The energy change in an endothermic reaction is: A. Internal B. External C. Negative D. Positive

77julia77 [94]

Answer:

Positive

Explanation:

In an endothermic reaction, the products are at a higher energy than the reactants. This means that the enthalpy change of the reaction (∆H) is positive

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3 years ago

A parachutist falls 50.0 m without friction. When the parachute opens, he slows down at a rate of 67 m/s*2. If he reaches the gr

KIM [24]

Answer:

3.49 seconds

3.75 seconds

-43200 ft/s²

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

$s=ut+\frac{1}{2}at^2\\\Rightarrow 50=0t+\frac{1}{2}\times 9.81\times t^2\\\Rightarrow t=\sqrt{\frac{50\times 2}{9.81}}\\\Rightarrow t=3.19\ s$

Time the parachutist falls without friction is 3.19 seconds

$v^2-u^2=2as\\\Rightarrow v=\sqrt{2as+u^2}\\\Rightarrow v=\sqrt{2\times 9.81\times 50+0^2}\\\Rightarrow v=31.32\ m/s$

Speed of the parachutist when he opens the parachute 31.32 m/s. Now, this will be considered as the initial velocity

$v=u+at\\\Rightarrow 11=31.32+9.81t\\\Rightarrow t=\frac{11-31.32}{-67}=0.3\ s$

So, time the parachutist stayed in the air was 3.19+0.3 = 3.49 seconds

$s=ut+\frac{1}{2}at^2\\\Rightarrow \frac{s}{2}=0t+\frac{1}{2}\times a\times t^2\\\Rightarrow \frac{s}{2}=\frac{1}{2}at^2$

$s=ut+\frac{1}{2}at^2\\\Rightarrow \frac{s}{2}=u1.1+\frac{1}{2}\times a\times 1.1^2$

Now the initial velocity of the last half height will be the final velocity of the first half height.

$v=u+at\\\Rightarrow v=at$

Since the height are equal

$\frac{1}{2}at^2=u1.1+\frac{1}{2}\times a\times 1.1^2\\\Rightarrow \frac{1}{2}at^2=at1.1+\frac{1}{2}\times a\times 1.1^2\\\Rightarrow 0.5t^2-1.1t-0.605=0\\\Rightarrow 500t^2-1100t-605=0$

$t=\frac{11\left(1+\sqrt{2}\right)}{10},\:t=\frac{11\left(1-\sqrt{2}\right)}{10}\\\Rightarrow t=2.65, -0.45$

Time taken to fall the first half is 2.65 seconds

Total time taken to fall is 2.65+1.1 = 3.75 seconds.

When an object is thrown with a velocity upwards then the velocity of the object at the point to where it was thrown becomes equal to the initial velocity.

$v^2-u^2=2as\\\Rightarrow a=\frac{v^2-u^2}{2s}\\\Rightarrow a=\frac{0^2-240^2}{2\times \frac{8}{12}}\\\Rightarrow a=-43200\ ft/s^2$

Magnitude of acceleration is -43200 ft/s²

5 0

3 years ago

A rocket sled accelerates from 10 m/s to 40 m/s in 2 seconds. What is the average acceleration of the sled?

Dafna11 [192]

Answer:

15m/s²

Explanation:

Given parameters:

Initial velocity = 10m/s

Final velocity = 40m/s

Time taken = 2s

Unknown:

Average acceleration = ?

Solution:

Acceleration is the rate of change of velocity with time;

Acceleration = $\frac{Final velocity - Initial velocity }{time}$

Acceleration = $\frac{40 - 10}{2}$ = 15m/s²

6 0

3 years ago

Two balls, ball A and ball B, are dropped from the same height onto the same surface. If ball A rebounds to a higher height than

mamaluj [8]

Answer:

its b

Explanation:

3 0

3 years ago

for any object suspended by any number of ropes, wires, or chains, how is the total amount of tension (tension in each rope adde

Sveta_85 [38]

Answer:

To calculate the tension on a rope holding 1 object, multiply the mass and gravitational acceleration of the object. If the object is experiencing any other acceleration, multiply that acceleration by the mass and add it to your first total.

Explanation:

The tension in a given strand of string or rope is a result of the forces pulling on the rope from either end. As a reminder, force = mass × acceleration. Assuming the rope is stretched tightly, any change in acceleration or mass in objects the rope is supporting will cause a change in tension in the rope. Don't forget the constant acceleration due to gravity - even if a system is at rest, its components are subject to this force. We can think of a tension in a given rope as T = (m × g) + (m × a), where "g" is the acceleration due to gravity of any objects the rope is supporting and "a" is any other acceleration on any objects the rope is supporting.[2]

For the purposes of most physics problems, we assume ideal strings - in other words, that our rope, cable, etc. is thin, massless, and can't be stretched or broken.

As an example, let's consider a system where a weight hangs from a wooden beam via a single rope (see picture). Neither the weight nor the rope are moving - the entire system is at rest. Because of this, we know that, for the weight to be held in equilibrium, the tension force must equal the force of gravity on the weight. In other words, Tension (Ft) = Force of gravity (Fg) = m × g.

Assuming a 10 kg weight, then, the tension force is 10 kg × 9.8 m/s2 = 98 Newtons.

7 0

3 years ago