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ella [17]
3 years ago
8

In a dark room you see objects when you shine a flashlight because

Physics
1 answer:
hammer [34]3 years ago
3 0
C illuminated by the flashlight. Same way that you see the moon ... it is illuminated by the sun.
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a person throws a ball in such a way that its speed is zero at one particular point in its path. How did the person throw the ba
Keith_Richards [23]

vertical! or in other words, up!


6 0
4 years ago
Can you please help me find the answer
Serggg [28]
Im sorry may you please retake the picture then i will answer
8 0
4 years ago
Can someone help me with this Physics question please?
strojnjashka [21]

Answer:

Explanation:

The formula for this, the easy one, is

N=N_0(\frac{1}{2})^{\frac{t}{H} where No is the initial amount of the element, t is the time in years, and H is the half life. Filling in:

N=48.0(\frac{1}{2})^{\frac{49.2}{12.3} and simplifying a bit:

N=48.0(.5)^4 and

N = 48.0(.0625) so

N = 3 mg left after 12.3 years

6 0
3 years ago
Read 2 more answers
You stand on a merry-go-round which is spinning at f = 0:25 revolutions per second. You are R = 200 cm from the center. (a) Find
ivanzaharov [21]

Answer:

(a) ω = 1.57 rad/s

(b) ac = 4.92 m/s²

(c) μs = 0.5

Explanation:

(a)

The angular speed of the merry go-round can be found as follows:

ω = 2πf

where,

ω = angular speed = ?

f = frequency = 0.25 rev/s

Therefore,

ω = (2π)(0.25 rev/s)

<u>ω = 1.57 rad/s </u>

(b)

The centripetal acceleration can be found as:

ac = v²/R

but,

v = Rω

Therefore,

ac = (Rω)²/R

ac = Rω²

therefore,

ac = (2 m)(1.57 rad/s)²

<u>ac = 4.92 m/s² </u>

(c)

In order to avoid slipping the centripetal force must not exceed the frictional force between shoes and floor:

Centripetal Force = Frictional Force

m*ac = μs*R = μs*W

m*ac = μs*mg

ac = μs*g

μs = ac/g

μs = (4.92 m/s²)/(9.8 m/s²)

<u>μs = 0.5</u>

7 0
3 years ago
An green hoop with mass mh = 2.8 kg and radius rh = 0.17 m hangs from a string that goes over a blue solid disk pulley with mass
vladimir2022 [97]
The mass of the hoop is the only force which is computed by:F net = 2.8kg*9.81m/s^2 = 27.468 N 
the slow masses that must be quicker are the pulley, ring, and the rolling sphere. 
The mass correspondent of M the pulley is computed by torque τ = F*R = I*α = I*a/R F = M*a = I*a/R^2 --> M = I/R^2 = 21/2*m*R^2/R^2 = 1/2*m 
The mass equal of the rolling sphere is computed by: the sphere revolves around the contact point with the table. So using the proposition of parallel axes, the moment of inertia of the sphere is I = 2/5*mR^2 for spin about the midpoint of mass + mR^2 for the distance of the axis of rotation from the center of mass of the sphere. I = 7/5*mR^2 M = 7/5*m 
the acceleration is then a = F/m = 27.468/(2.8 + 1/2*2 + 7/5*4) = 27.468/9.4 = 2.922 m/s^2
6 0
3 years ago
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