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Andrej [43]
4 years ago
15

I need answer really fast.

Physics
2 answers:
kumpel [21]4 years ago
8 0
I think it’s ( Only the arm length should be changed. The mass should remain the same )
sleet_krkn [62]4 years ago
3 0

Answer:

Only the arm length should be changed

Explanation:

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If a car accelerates uniformly from rest to 15 meters
Talja [164]

Answer:

1.125m/s^2

Explanation:

Since acceleration is defined as the rate of change in velocity with respect to time. Mathematically

v^2= u^2+2as

Where a,v,u and s are the acceleration, final velocity, initial velocity and distance respectively.

a = ?

u = 0m/s

v = 15m/s

s = 100m

Substituting the values into the formula above

v^2= u^2+2as

15^2=0^2+2×a×100

225= 0+200a

225= 200a

Divide both sides by 200

225/200 = 200a/200

a= 1.125m/s^2

Hence the acceleration of the car is 1.125m/s^2.

Note that the car accelerated uniformly from rest, that was why the initial velocity was 0m/s

8 0
3 years ago
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A girl hits a baseball with a bat. When the bat strikes the ball, what is the
Illusion [34]

Answer:

Its simple! B The ball accelerates upward!♡♡

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3 years ago
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A ________ is a government-regulated maximum price for goods.
Simora [160]
<span>D. price ceiling
</span><span>This is a government regulation that establishes a maximum price for a particular good.</span><span>

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8 0
3 years ago
Sunlight reflects from a concave piece of broken glass, converging to a point 34 cm from the glass. what is the radius of curvat
wolverine [178]
The rays of light coming from the Sun are parallel to each other, so when they are reflected by the concave piece of glass (which acts as a concave mirror) they converge into the focus of the mirror, which is
f=34 cm
The radius of curvature of a concave mirror is twice its focal length, so in this case it is:
r=2f = 2 \cdot 34 cm=78 cm

6 0
3 years ago
An investigator collects a sample of a radioactive isotope with an activity of 490,000 Bq.48 hours later, the activity is 110,00
Daniel [21]

Answer:

The correct answer is "22.27 hours".

Explanation:

Given that:

Radioactive isotope activity,

= 490,000 Bq

Activity,

= 110,000 Bq

Time,

= 48 hours

As we know,

⇒ A = A_0 e^{- \lambda t}

or,

⇒ \frac{A}{A_0}=e^{-\lambda t}

By taking "ln", we get

⇒ ln \frac{A}{A_0}=- \lambda t

By substituting the values, we get

⇒ -ln \frac{110000}{490000} = -48 \lambda

⇒    -1.4939=-48 \lambda

                 \lambda = 0.031122

As,

⇒ \lambda = \frac{ln_2}{\frac{T}{2} }

then,

⇒ \frac{ln_2}{T_ \frac{1}{2} } =0.031122

⇒ T_\frac{1}{2}=\frac{ln_2}{0.031122}

         =22.27 \ hours  

3 0
3 years ago
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