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natta225 [31]
3 years ago
5

A 0.5 kg air-track car is attached to the end of a horizontal spring of constant k = 20 N/m. The car is displaced 15 cm from its

equilibrium point and released.
a) What is the car's maximum speed? b) What is the car's maximum acceleration? c) What is the frequency f of the car's oscillation?
I need help please. What equations should I use for each problems?
Physics
1 answer:
Dimas [21]3 years ago
3 0
Find the amount of work that the spring does. This can be found using the equation 1/2kx^2. Then, you must set that equal to the amount of kinetic energy the car has. This is possible thanks to the work-energy theorem.

1/2kx^2 = 1/2mv^2

Solve to find velocity. Remember, the spring is displaced .15 m, not 15!

To find the acceleration, use F = ma. The force being applied to the car is kx, and you know the mass. You do the math.

For problem C I don't know, haven't done that yet in my class. Sorry!
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Two air track carts move along an air track towards each other. Cart A has a mass of 450 g and moves toward the right with a spe
ra1l [238]

Answer:

0.465 kgm/s

Explanation:

Given that

Mass of the cart A, m1 = 450 g

Speed of the cart A, v1 = 0.85 m/s

Mass of the cart B, m2 = 300 g

Speed of the cart B, v2 = 1.12 m/s

Now, using the law of conservation of momentum.

It is worthy of note that our cart B is moving in opposite directions to A

m1v1 + m2v2 =

(450 * 0.85) - (300 * 1.12) =

382.5 - 336 =

46.5 gm/s

If we convert to kg, we have

46.5 / 100 = 0.465 kgm/s

Thus, the total momentum of the system is 0.465 kgm/s

6 0
3 years ago
A pulley with the radius of 10.0 cm was connected to a motor with a massless
kogti [31]

Answer:

(i) -556 rad/s²

(ii) 17900 revolutions

(iii) 11250 meters

(iv) -55.6 m/s²

(v) 18 seconds

Explanation:

(i) Angular acceleration is change in angular velocity over time.

α = (ω − ω₀) / t

α = (10000 − 15000) / 9

α ≈ -556 rad/s²

(ii) Constant acceleration equation:

θ = θ₀ + ω₀ t + ½ αt²

θ = 0 + (15000) (9) + ½ (-556) (9)²

θ = 112500 radians

θ ≈ 17900 revolutions

(iii) Linear displacement equals radius times angular displacement:

s = rθ

s = (0.100 m) (112500 radians)

s = 11250 meters

(iv) Linear acceleration equals radius times angular acceleration:

a = rα

a = (0.100 m) (-556 rad/s²)

a = -55.6 m/s²

(v) Angular acceleration is change in angular velocity over time.

α = (ω − ω₀) / t

-556 = (0 − 15000) / t

t = 27

t − 9 = 18 seconds

8 0
3 years ago
A firewoman dropped a person onto the safety net. Right before the person hit the net he had a
sergejj [24]

Answer:

The Mass of a person is calculated to be 28.6kg .

Explanation:

This is based on Kinetic energy and we know, that Kinetic energy is the energy possessed by body by virtue of its motion .

It can be calculated by expression :

K.E=1/2mv²

Velocity of a person = 11.2m/sec

Kinetic energy = 1800J

Mass of person = ?

We know ,

K.E=1/2mv²

so, putting values we have :

1800=1/2 x m x (11.2)²

That is ,

m=1800 x 2 /11.2 x 11.2

or

m=3600/125.44

m = 28.6 kg

6 0
3 years ago
Two metal spheres of different sizes are charged such that the electric potential is the same at the surface of each. Sphere A h
vodka [1.7K]

Answer:

Explanation:

Given

Radius of A is twice of B i.e.

R_A=2R_B

Also Potential of both sphere is same

V_A=V_B

V=\frac{kQ}{R}

thus

k\frac{Q_A}{R_A}=K\frac{Q_B}{R_B}

\frac{Q_A}{Q_B}=\frac{R_A}{R_B}

\frac{Q_A}{Q_B}=\frac{2}{1}=2

\frac{Q_B}{Q_A}=\frac{1}{2}

(b)Ratio of \frac{E_B}{E_A}

Electric Field is given by E=\frac{kQ}{R^2}

thus E_A=\frac{kQ_A}{R_A^2}----1

E_B=\frac{kQ_B}{R_B^2}----2

Divide 2 by 1

\frac{E_B}{E_A}=\frac{Q_B}{R_B^2}\times \frac{R_A^2}{Q_A}

\frac{E_B}{E_A}=\frac{1}{2}\times 4=2

8 0
3 years ago
The diagram below shows the movement of matter in a portion of the water cycle.
Alisiya [41]

Answer:

d) precipitation

Hope it helps you

And if you want to, pls mark it as the brainliest answer

7 0
3 years ago
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