Magnetic pole reversal worksheet

A 7-kg bowling ball moving at 4 m/s strikes a 1 kg bowling pin. If the ball slows to 2 m/s in 0.05 s, how much force does it exe

iren [92.7K]

Answer:

280 N

Explanation:

acceleration = v2-v1 / time taken = (2-4 )/ 0.05 = -40 m/s^2 ( neg sign indicates slowing down )

force exerted = ma = 7 kg x -40 m/s^2 = - 280 N ( neg sign means opposite direction of initial velocity )

since the 7 kg ball is slowing down, the direction of force will be opposite of the initial velocity , and it will be 280 N

7 0

2 years ago

How does the intensity of a sound wave change if the distance from the

Alex17521 [72]

Answer:

The new intensity decreases by a factor of 16.

Explanation:

The intensity of sound wave is given by :

$I=\dfrac{P}{A}$

P is power

A is area

$I=\dfrac{P}{4\pi r^2}$

or

$I\propto \dfrac{1}{r^2}$ , r is distance from the source

If the distance from the source is increased by a factor of 4, r' = 4r

So,

$I'=\dfrac{1}{r'^2}\\\\I'=\dfrac{1}{(4r)^2}\\\\I'=\dfrac{1}{16}\times \dfrac{1}{r^2}\\\\I'=\dfrac{I}{16}$

So, the new intensity decreases by a factor of 16.

8 0

2 years ago

A person with a mass of 72 kg is riding a bicycle is accelerating at a rate of 5m/s on a horizontal surface. What is the weight

nikklg [1K]

Answer:

w = 706.32 [N]

Explanation:

The force due to gravitational acceleration can be calculated by means of the product of mass by gravitational acceleration.

w = m*g

where:

w = weight [N] (units of Newtons)

m = mass = 72 [kg]

g = gravity acceleration = 9.81 [m/s²]

Then we have:

$w = 72*9.81\\w = 706.32 [N]$

7 0

2 years ago

What is tan θ for the given triangle?

yulyashka [42]

Tan = opposite/adjacent
= 20/15
=4/3

7 0

2 years ago

Read 2 more answers

The hot glowing surfaces of stars emit energy in the form of electromagnetic radiation. It is a good approximation to assume tha

belka [17]

Given that,

Energy $H=2.7\times10^{31}\ W$

Surface temperature = 11000 K

Emissivity e =1

(a). We need to calculate the radius of the star

Using formula of energy

$H=Ae\sigma T^4$

$A=\dfrac{H}{e\sigma T^4}$

$4\pi R^2=\dfrac{H}{e\sigma T^4}$

$R^2=\dfrac{H}{e\sigma T^4\times4\pi}$

Put the value into the formula

$R=\sqrt{\dfrac{2.7\times10^{31}}{1\times5.67\times10^{-8}\times(11000)^4\times 4\pi}}$

$R=5.0\times10^{10}\ m$

(b). Given that,

Radiates energy $H=2.1\times10^{23}\ W$

Temperature T = 10000 K

We need to calculate the radius of the star

Using formula of radius

$R^2=\dfrac{H}{e\sigma T^4\times4\pi}$

Put the value into the formula

$R=\sqrt{\dfrac{2.1\times10^{23}}{1\times5.67\times10^{-8}\times(10000)^4\times4\pi}}$

$R=5.42\times10^{6}\ m$

Hence, (a). The radius of the star is $5.0\times10^{10}\ m$

(b). The radius of the star is $5.42\times10^{6}\ m$

8 0

2 years ago