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2 years ago

What would cause gravity’s force and why would it?

Physics

1 answer:

pickupchik [31]2 years ago

8 0

Answer:

Earth's gravity comes from all its mass. All its mass makes a combined gravitational pull on all the mass in your body. ... You exert the same gravitational force on Earth that it does on you. But because Earth is so much more massive than you, your force doesn't really have an effect on our planet.

Explanation:

You might be interested in

A boy pulls his 9.0 kg sled, applying a horizontal force of 14.0 N (rightward). The coefficient of friction between the snow and

Mama L [17]

Here is the answer, dud

5 0

2 years ago

Is diet Pepsi heterogeneous or is it homogeneous

Lana71 [14]

Heterogeneous because it has less amount of sugar than the regular Pepsi has

6 0

2 years ago

Read 2 more answers

Two strings with linear densities of 5 g/m are stretched over pulleys, adjusted to have vibrating lengths of 0.50 m, and attache

HACTEHA [7]

Answer:

2.18 kg

Explanation:

The frequency of a wave in a stretched string f = n/2L√(T/μ) where n = harmonic number, L = length of string, T = tension = mg where m = mass of object on string and g = acceleration due to gravity = 9.8 m/s² and μ = linear density of string.

For string 1, its fundamental frequency f is when n = 1. So,

f = 1/2L√(T/μ) = 1/2L√(mg/μ)

Now for string 1, L = 0.50 m, m = 20 kg and μ = 5 g/m = 0.005 kg/m

substituting the values of the variables into f, we have

f = 1/2L√(mg/μ)

f = 1/2 × 0.50 m√(20 kg × 9.8 m/s²/0.005 kg/m)

f = 1/1 m√(196 kgm/s²/0.005 kg/m)

f = 1/1 m√(39200 m²/s²)

f = 1/1 m × 197.99 m/s

f = 197.99 /s

f = 197.99 Hz

f ≅ 198 Hz

For string 2, at its third harmonic frequency f' is when n = 3. So,

f' = 3/2L√(T/μ) = 3/2L√(mg/μ)

Now for string 2, L = 0.50 m, m = M kg and μ = 5 g/m = 0.005 kg/m

substituting the values of the variables into f, we have

f' = 3/2L√(Mg/μ)

f' = 3/2 × 0.50 m√(M × 9.8 m/s²/0.005 kg/m)

f' = 3/1 m√(M1960 m²/s²kg)

f' = 3/1 m√M√(1960 m²/s²kg)

f' = 3/1 m √M × 44.27 m/s√kg

f' = 132.81√M/s√kg

f' = 132.81√M Hz/√kg

Since the frequency of the beat heard is 2 Hz,

f - f' = 2 Hz

So, 198 Hz - 132.81√M Hz/√kg = 2 Hz

132.81√M Hz/√kg = 198 Hz - 2 Hz

132.81√M Hz/√kg = 196 Hz

√M Hz/√kg = 196 Hz/138.81 Hz

√M/√kg = 1.476

squaring both sides,

[√M/√kg] = (1.476)²

M/kg = 2.178

M = 2.178 kg

M ≅ 2.18 kg

8 0

2 years ago

The floor of a railroad flatcar is loaded with loose crates having a coefficient of static friction of 0.420 with the floor. If

bonufazy [111]

Answer:

The distance is $s= 30.3 \ m$

Explanation:

From the question we are told that

The coefficient of static friction is $\mu_s = 0.42$

The initial speed of the train is $u = 57 \ km /hr = 15.8 \ m/s$

For the crate not to slide the friction force must be equal to the force acting on the train i.e

$-F_f = F$

The negative sign shows that the two forces are acting in opposite direction

=> $mg * \mu_s = ma$

=> $-g * \mu_s = a$

=> $a = -9.8 * 0.420$

=> $a = -4.116 m/s^2$

From equation of motion

$v^2 = u^2 + 2as$

Here v = 0 m/s since it came to a stop

=> $s= \frac{v^2 - u^2 }{ 2 a}$

=> $s= \frac{0 -(15.8)^2 }{ - 2 * 4.116}$

=> $s= 30.3 \ m$

7 0

2 years ago

You have just moved into a new apartment and are trying to arrange your bedroom. You would like to move your dresser of weight 3

LenaWriter [7]

Answer:

0 Joules

Explanation:

The work done is given by

$W=F\times s\times cos\theta$

where,

F = Force applied

s = Displacement of the object = 0 m

$\theta$ = Angle between the force applied and the horizontal = 0

$W=F\times 0\times cos0\\\Rightarrow W=0\ J$

Work is only observed when there is a displacement.

The work done by me is 0 Joules as I was unable to move it.

6 0

2 years ago