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lana [24]
3 years ago
13

A 15 kg runaway grocery cart runs into a spring with spring constant 230 N/m and compresses it by 56 cm .What was the speed of t

he cart just before it hit the spring?
Physics
1 answer:
liubo4ka [24]3 years ago
3 0

To solve this problem we will apply the concepts related to the conservation of kinetic energy and elastic potential energy. Thus we will have that the kinetic energy is

KE = \frac{1}{2} mv^2

And the potential energy is

PE = \frac{1}{2} kx^2

Here,

m = mass

v = Velocity

x = Displacement

k = Spring constant

There is equilibrium, then,

KE = PE

\frac{1}{2} mv^2 = \frac{1}{2} kx^2

Our values are given as,

x=0.56m\\k=230N/m\\m=15kg

Replacing we have that

\frac{1}{2} (15)v^2 = \frac{1}{2} (230)(0.56)^2

v = 2.19m/s

Therefore the speed of the cart is 2.19m/s

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A kangaroo jumps straight up to a vertical height of 1.45 m. How long was it in the air before returning to Earth?
dexar [7]

Answer:

1.08 s

Explanation:

From the question given above, the following data were obtained:

Height (h) reached = 1.45 m

Time of flight (T) =?

Next, we shall determine the time taken for the kangaroo to return from the height of 1.45 m. This can be obtained as follow:

Height (h) = 1.45 m

Acceleration due to gravity (g) = 9.8 m/s²

Time (t) =?

h = ½gt²

1.45 = ½ × 9.8 × t²

1.45 = 4.9 × t²

Divide both side by 4.9

t² = 1.45/4.9

Take the square root of both side

t = √(1.45/4.9)

t = 0.54 s

Note: the time taken to fall from the height(1.45m) is the same as the time taken for the kangaroo to get to the height(1.45 m).

Finally, we shall determine the total time spent by the kangaroo before returning to the earth. This can be obtained as follow:

Time (t) taken to reach the height = 0.54 s

Time of flight (T) =?

T = 2t

T = 2 × 0.54

T = 1.08 s

Therefore, it will take the kangaroo 1.08 s to return to the earth.

3 0
3 years ago
A jogger hears a car alarm and decides to investigate. While running toward the car, she hears an alarm frequency of 872.10 Hz.
Nata [24]

Answer:

v = 4.18 m/s

Explanation:

given,

frequency of the alarm = 872.10 Hz

after passing car frequency she hear = 851.10 Hz

Speed of sound = 343 m/s

speed of the jogger = ?

speed of the

v_f = \dfrac{872.10-851.10}{2}

v_f =10.5\ Hz

v_o = 872.1 - 10.5

V_0 = 861.6\ Hz

The speed of jogger

v = \dfrac{v_1 \times 343}{v_0}-343

v = \dfrac{872.1 \times 343}{861.6}-343

v = 4.18 m/s

5 0
3 years ago
A mass on a horizontal surface is connected to the spring and pulled to the right along the surface stretching the spring by 25
solniwko [45]

Answer:

320 N/m

Explanation:

From Hooke's law, we deduce that

F=kx where F is applied force, k is spring constant and x is extension or compression of spring

Making k the subject of formula then

k=\frac {F}{x}

Conversion

1m equals to 100cm

Xm equals 25 cm

25/100=0.25 m

Substituting 80 N for F and 0.25m for x then

k=\frac {80}{0.25}=320N/m

Therefore, the spring constant is equal to 320 N/m

3 0
4 years ago
If the object represented by the FBD below has a mass of 2.5 kg, what is the acceleration of the object?
Debora [2.8K]

Answer:

4 m/s² down

Explanation:

We'll begin by calculating the net force acting on the object.

The net force acting on the object from the left and right side is zero because the same force is applied on both sides.

Next, we shall determine the net force acting on the object from the up and down side. This can be obtained as follow:

Force up (Fᵤ) = 15 N

Force down (Fₔ) = 25 N

Net force (Fₙ) =?

Fₙ = Fₔ – Fᵤ

Fₙ = 25 – 15

Fₙ = 10 N down

Finally, we shall determine the acceleration of the object. This can be obtained as follow:

Mass (ml= 2.5 Kg

Net force (Fₙ) = 10 N down

Acceleration (a) =?

Fₙ = ma

10 = 2.5 × a

Divide both side by 2.5

a = 10 / 2.5

a = 4 m/s² down

Therefore, the acceleration of the object is 4 m/s² down

6 0
3 years ago
1. When in the past have you pushed your personal limits? Give at least one
Yuri [45]

Answer:

Umm that's a personal question. All u have to do is say when have u pushed your personal limits....... Ummm one for me is when i had to try out for a select soccer and that is past my comfort zone.  

Explanation:

4 0
3 years ago
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