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3 years ago

Can somebody please help?

Physics

1 answer:

DedPeter [7]3 years ago

7 0

Answer:

Im not sure

Explanation:

I don't take physics cuz im in 9th grade. so. idk but I will find out and come back with an answer.

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Why is it wrong to leave our light on

Dennis_Churaev [7]

Answer:

you will get huge electricity bills ............

8 0

3 years ago

Two cars start from rest at a red stop light. When the light turns green, both cars accelerate forward. The blue car accelerates

saul85 [17]

First, create an illustration of the motion of the two cars as shown in the attached picture. The essential equations used is

For constant acceleration:
a = v,final - v,initial /t

The solutions is as follows:

a = v,final - v,initial /t
3.8 = (v - 0)/2.8 s
v = 10.64 m/s
After 2.8 seconds, the speed of the blue car is 10.64 m/s.

4 0

3 years ago

A diver makes 1.0 revolutions on the way from a 9.2-m-high platform to the water. Assuming zero initial vertical velocity, find

____ [38]

Answer:

Average angular velocity ≈ 4.59 rad/s

Explanation:

Using the equation of motion,

H = ut + (1/2)t² ............................ equation 1.

Where H= height, u = initial velocity(m/s), g = acceleration due to gravity(m/s²), t = time(s) u= 0 ∴ ut =0

H =(1/2)gt².................................... equation 2.

making t² the subject of the relation in equation 2,

∴ t² = 2H/g

Where H = 9.2 m, g= 9.8 m/s

∴ t² = ( 2×9.2)/9.8

t = √(2 × 9.2/9.8) = √(18.4/9.8)

t = 1.37 s.

The average angular velocity = θ/t

Where θ = is the number of revolution that the diver makes, t = time

θ = 1 rev.

Since 1 rev = 2π (rad)

t = 1.37 s

Average angular velocity = 2π/t

π = 3.143

Average angular velocity = (2×3.143)/1.37 = 6.286/1.37

Average angular velocity ≈ 4.59 rad/s

8 0

3 years ago

If two point masses 1kg & 4kg are seperated by a distance of 2m. Magnitude of gravitational force exerted by 1kg on 4kg is ?

Aliun [14]

Answer:

F = G Newtons

Explanation:

Given:

Mass of 1st body = $1\:kg$
Mass of 2nd body = $4\:kg$

To Find:

Magnitude of gravitational force

Solution:

Here, we have a formula

$F=\dfrac{G.M_{1}.M_{2}}{r^{2}}$

<u>Substituting the values</u>

$\implies\:\:F = \dfrac{G(1)(4)}{2^{2}}$

$\implies\:\:F = \dfrac{4G}{4}$

$\implies\:\:F = \dfrac{\cancel{4}G}{\cancel{4}}$

$\implies\:\:\red{F = G}$

Know More:

The applied formula for the above solution is

${\boxed{F_{G}=\dfrac{G.M_{1}.M_{2}}{r^{2}}}}$

where,

F $_{G}$ = Gravitational force
G = Gravitational constant
M $_{1}$ = mass of 1st body
M $_{2}$ = mass of 2nd body
r = distance between two bodies

6 0

3 years ago

A piece of amber is charged by rubbing with a piece of fur. If the net excess charge on the fur is 8.9 nC ( 8.9 10-9 C), how man

Alika [10]

Answer:

Number of electrons, $n=5.56\times 10^{10}$

Explanation:

Given that,

Charge on the fur, $q=8.9\ nC=8.9\times 10^{-9}\ C$

A piece of amber is charged by rubbing with a piece of fur. We need to find the electrons were added to the amber. It can be calculated using the quantization of charge as

q = n × e

Where

n is the number of electrons

e is the charge on electron

$n=\dfrac{q}{e}$

$n=\dfrac{8.9\times 10^{-9}}{1.6\times 10^{-19}}$

$n=5.56\times 10^{10}\ electrons$

So, $n=5.56\times 10^{10}\ electrons$ number of electrons are added to the amber. Therefore, this is the required solution.

7 0

4 years ago