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Anastasy [175]
2 years ago
10

Block A is accelerating with Block B at a rate of 0.800 m/s2 along a frictionless surface. It suddenly encounters a surface that

supplies 25.0 N a friction. What is the new Acceleration?
A. 0.4 m/s2

B. 1.0 m/s2

C. 0.8 m/s2

D. 0.6 m/s2

Physics
1 answer:
Natalija [7]2 years ago
7 0

Answer:

a=0.6\ m/s^2

Explanation:

The attached figure shows the whole description. Considering the applied force is 100 N.

The acceleration of both blocks A and B, a=0.8\ m/s^2

Firstly calculating the mass m using the second law of motion as :

F = ma

m is the mass

m=\dfrac{F}{a}

m=\dfrac{100\ N}{0.8\ m/s^2}

m = 125 kg

It suddenly encounters a surface that supplies 25.0 N a friction, F' = 25 N

(F-F')=ma

(100-25)=125\times a

a=\dfrac{75}{125}=0.6\ m/s^2

So, the new acceleration of the block is 0.6\ m/s^2. Hence, this is the required solution.

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Answer:

\boxed {\boxed {\sf 500 \ Newtons }}

Explanation:

The equation for force is given:

F=m*a

First, we must find the total mass, which is the sum of the boy's mass and the go-cart's mass.

  • total mass= boy's mass + go cart's mass

The boy's mass is 35 kilograms and the go cart's is 65 kilograms.

  • total mass= 35 kg+ 65 kg=100 kg

Now we know the total mass and the acceleration.

m= 100 \ kg \\a= 5 \ m/s^2

Substitute the values into the formula.

F=100 \ kg * 5 \ m/s^2

Multiply.

F= 500 \ kg*m/s^2

  • 1 kilograms meter per square second is equal to 1 Newton.
  • Our answer of 500 kg*m/s² is equal to 500 Newtons.

F= 500 \ N

The force exerted by the go cart engine is <u>500 Newtons.</u>

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2 years ago
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Oxygen-18 is a naturally-occuring, stable isotope and is commonly used is scientific studies as a tracer. Using the periodic tab
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Oxygen has Atomic number 8 so all isotopes have 8 protons and 8 electrons.

So the number of neutrons in Oxygen-18 = 18 - 8 = 10.

Option B is the correct one.


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3 years ago
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A racquetball with a mass of 42 g is moving with a horizontal speed of 7 m/s to the right (+x direction). It hits the wall of th
zheka24 [161]

The magnitude of the racquetball's change in momentum is 0.59 kgm/s approximately.

Given that a racquetball with a mass of 42 g is moving with a horizontal speed of 7 m/s to the right (+x direction).

mass m  = 42g = 42/1000 = 0.042kg

initial velocity before collision u = 7 m/s

It hits the wall of the court and rebounds to the hitter with a horizontal speed of 7m/s to the left (-x direction). That is,

velocity after collision v = 7 m/s

To calculate the magnitude of the racquetball's change in momentum, we will use the formula below

Change in momentum = Mv - Mu

Since momentum is a vector quantity, we will consider the direction.

Change in momentum = 0.042 x 7 - ( 0.042 x - 7)

Change in momentum = 0.294 + 0.294

Change in momentum = 0.588 kgm/s

Therefore, the magnitude of the racquetball's change in momentum is 0.59 kgm/s approximately.

Learn more on momentum here: brainly.com/question/402617

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2 years ago
Look at the circuit diagram. Which of these components is part of the circuit?
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2 years ago
A stone is dropped into a river from a bridge 41.7 m above the water. Another stone is thrown vertically down 1.80 s after the f
hram777 [196]

Answer:

31.75 m/s

Explanation:

h = 41.7 m

Let the initial velocity of the second stone is u

Let the time taken to reach to the bottom by the first stone is t then the time taken by the second stone to reach the ground is t - 1.8.

For first stone:

Use second equation of motion

h=ut+\frac{1}{2}gt^2

Here, u = 0, g = 9.8 m/s^2 and t be the time and h = 41.7

So, 41.7= 0 + 0.5 x 9.8 x t^2

41.7 = 4.9 t^2

t = 2.92 s ..... (1)

For second stone:

Use second equation of motion

h=ut+\frac{1}{2}gt^2

Here, g = 9.8 m/s^2 and time taken is t - 1.8 = 2.92 - 1.8 = 1.12 s, h = 41.7 m and u be the initial velocity

h=u\left ( t-1.8 \right )+4.9\left ( t-1.8 \right )^2    .... (2)

By equation the equation (1) and (2), we get

41.7=1.12 u +4.9 \times 1.12^{2}

u = 31.75 m/s

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3 years ago
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