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3 years ago

9

Driving at 75m/s, you hit the brakes on your new car, accelerating at a rate of -20m/s. How long will it take the car to come to

rest?

1 answer:

soldi70 [24.7K]3 years ago

5 0

Answer:

3.75s

Explanation:

Using equation of motion, t= (v-u)/a

where v=0, u=75 and a=-20

t=-75/-20=75/20=3 /.75

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A tennis ball, 0.314kg, is accelerated at a rate of 164m/s2 when hit by a professional tennis player. What force does the player

Colt1911 [192]

Newton's 2nd law of motion:

Force = (mass) x (acceleration)

= (0.314 kg) x (164 m/s²)

= 51.5 newtons

(about 11.6 pounds).

Notice that the ball is only accelerating while it's in contact with the racket. The instant the ball loses contact with the racket, it stops accelerating, and sails off in a straight line at whatever speed it had when it left the strings.

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3 years ago

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3 years ago

1. Bone has a Young’s modulus of about

Blababa [14]

#1

As we know that

$Y = \frac{stress}{strain}$

now plug in all data into this

$1.8\times 10^{10} = \frac{1.68 \times 10^8}{strain}$

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now from the formula of strain

$strain = \frac{\Delta L}{L}$

$9.33 \times 10^{-3} = \frac{\Delta L}{0.54}$

$\Delta L = 5.04 \times 10^{-3} m$

$\Delta L = 5.04 mm$

#2

As we know that

pressure * area = Force

here we know that

$Area = 3.53 \times 11.6 = 40.95 m^2$

$P = 0.2 atm = 0.2 \times 1.01 \times 10^5 = 2.02\times 10^4 Pa$

now force is given as

$F = 40.95 \times (2.02\times 10^4) = 8.27 \times 10^5 N$

#3

As we know that density of water will vary with the height as given below

$\rho = \frac{\rho_0}{1 - \frac{\Delta P}{B}}$

here we know that

$\Delta P = 2600 atm = 2600 \times 1.01 \times 10^5 = 2.63\times 10^8 Pa$

$B = 2.3 \times 10^9 N/m^2$

now density is given as

$\rho = \frac{1050}{1 - \frac{2.63\times 10^8}{2.3 \times 10^9}}$

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#4

as we know that pressure changes with depth as per following equation

$P = P_o + \rho g h$

here we know that

$P = 3 P_0$

now we will have

$3P_0 = P_0 + \rho g h$

$2P_0 = \rho g h$

$2(1.01 \times 10^5) = 1025 (9.81)(h)$

here we will have

$h = 20.1 m$

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#5

Here net buoyancy force due to water and oil will balance the weight of the block

so here we will have

$mg = \rho_1V_1g + \rho_2V_2g$

$A(0.0476)979 = 922(A)(0.0476 - x) + 1000(A)(x)$

$46.6 = 43.89 - 922x + 1000x$

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