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Dvinal [7]
3 years ago
11

The water skier has a mass of 73 kg. Find the magnitude of the

Physics
1 answer:
Papessa [141]3 years ago
6 0

To solve this problem we will apply the linear motion kinematic equations, specifically the concept of acceleration as a function of speed and time, as well as Newton's second law.

PART A) Acceleration can be described as changing the speed in a period of time therefore,

a = \frac{V}{t}\\a = \frac{11}{8}\\a = 1.375m/s^2

Force is the proportional change between mass and acceleration therefore

F = (73)(1.375)

F= 100.375N

PART B) We will apply the same concept given but now we will change the time to 21s therefore:

a = \frac{V}{t}\\a = \frac{11}{21}\\a = 0.5238

Now the force

F = (73)(0.5238)

F = 38.23N

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What do we call a substance in<br> which two or more elements are<br> chemically bonded
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A compound

Explanation:

A compound is a substance formed when two or more elements are chemically joined

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How can velocity change even if speed stays the same?​
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the change of direction it's going

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Please Help! When a metal is heated, the free electrons gain _____________.
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It’s B potential energy
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if a car that is 100 feet in front of you on route 70 west slams on the brakes while you are traveling 65 miles per hour (95 fee
fenix001 [56]

Answer:

t = 1.05 s

Explanation:

Given,

The distance between your vehicle and car, 100 ft

The constant speed of your vehicle, u = 95 ft/s

Since, the velocity is constant, a =0

If the car stopped suddenly, time left for you to hit the brake, t = ?

Using the second equation of motion,

                           S = ut + ½ at²

Substituting the given values in the equation

                           100 = 95 x t

                             t = 100/95

                               = 1.05 s

Hence, the time left for you to hit the brakes and stop before rear ending them, t = 1.05 s

5 0
2 years ago
A block is attached to a horizontal spring and oscillates back and forth on a frictionless horizontal surface at a frequency of
Nastasia [14]

Answer:

Amplitude = 0.058m

Frequency = 6.25Hz

Explanation:

Given

Amplitude (A) = 8.26 x 10-2 m

Frequency (f) = 4.42Hz

Conversation of energy before split

½mv² = ½KA²

Make A the subject of formula

A = v\sqrt{\frac{m}{k} }

Conversation of energy after split

½(m/2)V'² = ½(m/2)V² = ½KA'²

½(m/2)V² = ½KA'²

Make A the subject of formula

First divide both sides by ½

(m/2)V² = KA'²

Divide both sides by K

\frac{m}{2K}V² = A'²

v\sqrt{\frac{m}{2k} } = A'

Substitute v\sqrt{\frac{m}{k} } for A in the above equation

A' = A/√2

A' = 8.26 x 10^-2/√2

A' = 0.05840702012600882

Amplitude after split = 0.058 (Approximated)

Frequency (f') = f√2

f' = 4.42√2

f' = 6.25082394568908011

Frequency after split = 6.25Hz (approximated)

8 0
3 years ago
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