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Maurinko [17]
3 years ago
7

A car accelerates from rest. It reaches a velocity of 25m/s in 10 seconds. What was the cars acceleration?

Physics
1 answer:
mixer [17]3 years ago
3 0

Answer:

We are Given:

initial velocity (u) = 0m/s

final velocity (v) = 25 m/s

time (t) = 10 seconds

acceleration (a)  = a m/s/s

From the first equation of motion, we know that:

v = u + at

solving for a, we get:

a = (v-u) / t

now, plugging the given values in this equation

a = (25 - 0) / 10

a = 25 / 10

a = 2.5 m/s/s

Therefore, the acceleration of the car is 2.5 m/s/s

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Is the intensity of light directly or inversely proportional to the concentration according to Beer-lambert's law?
Ahat [919]

Answer:

directly proportional

Explanation:

The Beer-Lambert law states that the quantity of light absorbed by a substance dissolved in a fully transmitting solvent is directly proportional to the concentration of the substance and the path length of the light through the solution.

8 0
3 years ago
Which example possesses mechanical potential energy?. A. a taut guitar string. B. an oscillating pendulum. C. a roller coaster r
NeTakaya
<span>
The taut guitar string haspotencial energy which we can see in action.</span>  <span>· so option a is correct.</span>
7 0
3 years ago
Read 2 more answers
A(n) 60.3 g ball is dropped from a height of 53.7 cm above a spring of negligible mass. The ball compresses the spring to a maxi
Feliz [49]

Answer:

271.862 N/m

Explanation:

From Hook's Law,

mgh = 1/2ke²............... Equation 1

Where

m = mass of the ball, g = acceleration due to gravity, k = spring constant, e = extension, h = height fro which the ball was dropped.

Making k the subject of the equation,

k =2mgh/k²....................... Equation 2

Note: The potential energy of the ball is equal to the elastic potential energy of the spring.

Given: m = 60.3 g = 0.0603 kg, g = 9.8 m/s², e = 4.68317 cm = 0.0468317 m, h = 53.7 cm = 0.537 m

Substitute into equation 2

k = 2(0.0603)(9.8)(0.537)/0.048317²

k = 0.6346696/0.0023345

k = 271.862 N/m

7 0
3 years ago
A block of mass 200g is oscillating on the end of a horizontal spring of spring constant 100 N/m and natural length 12 cm. When
malfutka [58]

In order to determine the acceleration of the block, use the following formula:

F=ma

Moreover, remind that for an object attached to a spring the magnitude of the force acting over a mass is given by:

F=kx

Then, you have:

ma=kx

by solving for a, you obtain:

a=\frac{kx}{m}

In this case, you have:

k: spring constant = 100N/m

m: mass of the block = 200g = 0.2kg

x: distance related to the equilibrium position = 14cm - 12cm = 2cm = 0.02m

Replace the previous values of the parameters into the expression for a:

a=\frac{(\frac{100N}{m})(0.02m)}{0.2\operatorname{kg}}=10\frac{m}{s^2}

Hence, the acceleration of the block is 10 m/s^2

8 0
1 year ago
A ball dropped from a bridge takes three seconds to reach the water below how far is the bridge above the water?
tatuchka [14]

<u>Given that:</u>

Ball dropped from a bridge at the rate of 3 seconds

Determine the height of fall (S) = ?

      As we know that, S = ut + 1/2 ×a.t²

                          u =initial velocity = 0

                          a= g =9.81 m/s  (since free fall)

                            S = 0+ 1/2 × 9.81 × 3²

                          <em> S = 44.145 m</em>

<em>44.145 m far is the bridge from water</em>

6 0
3 years ago
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