All categories

2 years ago

If a 2 kg object is falling at 3 m/s at what rate is gravity working on the object

Physics

1 answer:

777dan777 [17]2 years ago

4 0

Answer:

+9.8m/s^2

Explanation:

The rate of gravity of the object is constant thriughout the surface of the earth.

For falling object, the rate of gravity is positive since the body is coming down (falling)

The rate of gravity is negative if the body is going up

The constant value for acceleration due to gravity is 9.8m.s^2

Since the object is falling, hence the acceleration due to gravity is positive.

Rate of gravity working on the object will be +9.8m/s^2

You might be interested in

Which was least likely to have been a component of Earth’s atmosphere before life began?

Otrada [13]

The answer to this question is a) sulfur

4 0

2 years ago

Read 2 more answers

A spherical helium filled balloon (B) with a hanging passenger cage being held by a single vertical cable (C) attached to Earth

Gre4nikov [31]

Answer:

The tension is $T = 4326.7 \ N$

Explanation:

From the question we are told that

The total mass is $m = 200 \ kg$

The radius is $r = 5 \ m$

The density of air is $\rho_a = 1.225 \ kg/m^3$

Generally the upward force acting on the balloon is mathematically represented as

$F_N = T + mg$

=> $(\rho_a * V * g ) = T + mg$

=> $T = (\rho_a * V * g ) - mg$

Here V is the volume of the spherical helium filled balloon which is mathematically represented as

$V = \frac{4}{3} * \pi r^3$

=> $V = \frac{4}{3} * 3.142 *(5)^3$

=> $V = 523.67\ m^3$

$T = (1.225 * 523.67* 9.8 ) - 200 * 9.8$

$T = 4326.7 \ N$

5 0

2 years ago

How are thunderstorms kept alive?

Burka [1]

Answer:

E step leader

Explanation:

5 0

2 years ago

Calculate the average speed of a runner who runs to for 500 meters in 40 second

siniylev [52]

Answer:

12.5

Explanation:

5 0

2 years ago

A projectile is shot at an angle 45 degrees to the horizontalnear the surface of the earth but in the absence of air resistance.

juin [17]

Answer:

Explanation:

Given

For first case

launch angle $\theta =45^{\circ}C$

at highest point $h=150 m/s$

$150=u\cos 45$

$u=\frac{150}{\cos 45}=212.132 m/s$

For second case

$\theta _2=37^{\circ}C$

at highest Point velocity is $u\cos \theta _2$

$=212.132\times \cos 37$

$=169.41 m/s$

as there is no acceleration in x direction therefore horizontal velocity is same

7 0

2 years ago

Read 2 more answers