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emmasim [6.3K]
3 years ago
11

A police car waits in hiding slightly off the highway. A speeding car is spotted by the police car doing 40 m/s. At the instant

the speeding car passes the police car, the police car accelerates from rest at 4 m/s2 to catch the speeding car. How long does it take the police car to catch the speeding car
Physics
1 answer:
eimsori [14]3 years ago
3 0

Answer:

t = 20s

Explanation:

In order for the police car to capture the other car, both cars must travel the same length.

The distance traveled by the first vehicle is:

S = ^ Vt = 40t, where t represents the elapsed time and V the vehicle speed.

The distance traveled by the police car is:

S = at ^ 2/2 = 2t ^ 2, where t represents the elapsed time, a is the acceleration of the car.

As the distances traveled are equal:

40t = 2t ^ 2 in this way t = 20s

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An air-track glider attached to a spring oscillates between the 10 cm mark and the 60 cm mark on the track. The glider completes
Assoli18 [71]

Answer:

a) Time period is 3.3 seconds

b) The frequency is 0.3030 Hz

c) amplitude is 0.25 m

d) maximum speed is 0.476 m/s

Explanation:

Given the data in the question;

a) Period

Time Period T = Time taken for one oscillation

T = 33s / 10 = 3.3 seconds

Therefore, Time period is 3.3 seconds

b) Frequency

we know that frequency is the inverse of time period

so;

Frequency f = 1/T = 1 / 3.3 s

Frequency f = 0.3030 Hz

Therefore, The frequency is 0.3030 Hz

c) amplitude

amplitude A = \frac{1}{2}( 60 cm - 10 cm )

A = \frac{1}{2} × 50 cm

A = 25 cm

A = 0.25 m

Therefore, amplitude is 0.25 m

d) maximum speed of the glider

maximum speed V_{max} = ωA

and ω = 2π/T

so maximum speed V_{max} = \frac{2\pi }{T}A

so we substitute

so maximum speed V_{max} = \frac{2\pi }{3.3} × 0.25 m

so maximum speed V_{max} = 0.476 m/s

Therefore, maximum speed is 0.476 m/s

5 0
3 years ago
If a 10kg block is at rest on a table and a 1200N force is applied in the eastward direction for 10 seconds, what is the acceler
gavmur [86]

Answer:

a = 120 m/s²

Explanation:

We apply Newton's second law in the x direction:

∑Fₓ = m*a Formula (1)

Known data

Where:

∑Fₓ: Algebraic sum of forces in the x direction

F: Force in Newtons (N)

m: mass (kg)

a: acceleration of the block (m/s²)

F = 1200N

m = 10 kg

Problem development

We replace the known data in formula (1)

1200 = 10*a

a = 1200/10

a = 120 m/s²

6 0
3 years ago
A runner exerts a net force of 335 n to accelerate at a rate of 2.5 m/s what is the runners mass
Olegator [25]
F=ma
M=F/a
M=335/2.5
M=134 kg
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