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djverab [1.8K]
3 years ago
10

It takes 160 kj of work to accelerate a car from 24.0 m/s to 27.5 m/s. what is the car's mass?

Physics
1 answer:
ankoles [38]3 years ago
4 0
Work-Energy :W = 1/2 m ( Vf^2 -Vo^2 )
Vo = 24.0 m/s Initial speed 
 Vf = 27.5 m/s  Final speed 

W = 1/2 m ( Vf^2 -Vo^2 )
160 kj = 1/ 2 m ( 27.5^2  -24.0 ^2)
160kj =  4680 x m
convert kilo joules to jeoules                     160000 j = 4689 xm
m = 160000 j/4689
m = 34.18 kg
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Unless you have a diagram to include or any other additional info, I'll assume the block is being pulled by two opposing forces along the horizontal surface.

Horizontally, the block is under the influence of

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It stands to reason that the block is accelerating in the direction of the larger pulling force.

(A) By Newton's second law, we have

15 N + (-5 N) + (-3 N) = <em>m</em> (1 m/s²)

where <em>m</em> is the mass of the block. Solve for <em>m</em> :

7 N = <em>m</em> (1 m/s²)

<em>m</em> = (7 N) / (1 m/s²)

<em>m</em> = 7 kg

(B) The friction force is proportional to the normal force, so that if <em>f</em> is the mag. of friction and <em>n</em> is the mag. of the normal force, then <em>f</em> = <em>µ</em> <em>n</em> where <em>µ</em> is the coefficient of friction.

The block does not bounce up and down, so its vertical forces are balanced, which means the normal force and the block's weight (mag. <em>w</em>) cancel out:

<em>n</em> + (-<em>w</em>) = 0

<em>n</em> = <em>w</em>

<em>n</em> = <em>m</em> <em>g</em>

where <em>g</em> = 9.8 m/s² is the mag. of the acceleration due to gravity.

<em>n</em> = (7 kg) (9.8 m/s²)

<em>n</em> = 68.6 N

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3 N = <em>µ</em> (68.6 N)

<em>µ</em> = (3 N) / (68.6 N)

<em>µ</em> ≈ 0.044

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Lonnie pitches a baseball of mass 0.20 kg. The ball arrives at home plate with a speed of 40 m/s and is batted straight back to
abruzzese [7]

Answer:

B) 20N.s is the correct answer

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Thus, we have

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Answer:

The time it takes the proton to return to the horizontal plane is 7.83 X10⁻⁷ s

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F/m = Eq/m = mg/m, then

g = Eq/m --------equation 1

Again, in a projectile motion, the time of flight (T) is given as

T = (2usinθ/g) ---------equation 2

Substitute in the value of g into equation 2

T = \frac{2usin \theta}{\frac{Eq}{m}} =\frac{m* 2usin \theta}{Eq}

Charge of proton = 1.6 X 10⁻¹⁹ C

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E is given as 400 N/C, u = 3.0 × 10⁴ m/s and θ = 30°

Solving for T;

T = \frac{(1.67X10^{-27}* 2*3X10^4sin 30}{400*1.6X10^{-19}}

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