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astraxan [27]
3 years ago
13

Plz help me it is improtant

Physics
1 answer:
marishachu [46]3 years ago
7 0
I think it is b cause I don’t think you do that
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a mass on a spring vibrates in simple harmonic motion at an amplitude of 8.0 cm. if the mass of the object is 0.20kg and the spr
Reil [10]

Answer:

4.06 Hz

Explanation:

For simple harmonic motion, frequency is given by

f=\frac {1}{2\pi}\times \sqrt{\frac {k}{m}} where k is spring constant and m is the mass of the object.

Substituting 0.2 Kg for mass and 130 N/m for k then

f=\frac {1}{2\pi}\times \sqrt{\frac {130}{0.2}}=4.057670803\\f\approx 4.06 Hz

5 0
3 years ago
A car with a mass of 1.50x10^3 kg starts from rest and accelerates to a speed of 18.0m/s in 12.0 s. assume that the force of res
Luden [163]
The first thing you should know for this case is the definition of distance.
 d = v * t
 Where,
 v = speed
 t = time
 We have then:
 d = v * t
 d = 9 * 12 = 108 m
 The kinetic energy is:
 K = ½mv²
 Where,
 m: mass
 v: speed
 K = ½ * 1500 * (18) ² = 2.43 * 10 ^ 5 J
 The work due to friction is
 w = F * d
 Where,
 F = Force
 d = distance:
 w = 400 * 108 = 4.32 * 10 ^ 4
 The power will be:
 P = (K + work) / t
 Where,
 t: time
 P = 2.86 * 10 ^ 5/12 = 23.9 kW
 answer:
 the average power developed by the engine is 23.9 kW
8 0
3 years ago
An 100W light bulb is on 6 hours work out the energy (kWh) used and the cost of it (1kWh = £0.1359) with working out please
Ainat [17]
1,000 W  =  1 kW
100 W  =  0.1 kW

           (0.1 kW) x (6 h) = 0.6 kWh    <=== energy

           (0.6 kWh) x (£0.1359/kWh)  =  £0.0815     <=== cost of it     
8 0
3 years ago
A silver tea spoon is placed in a cup filled with hot tea. After some time, the exposed end of the spoon becomes hot even withou
EastWind [94]

Answer:

As atoms in the spoon vibrates about their equilibrium positions and transfer energy form one end to other end. This process is called conduction.

4 0
3 years ago
A ball is dropped from rest from a height h above the ground. another ball is thrown vertically upwards from the ground at the i
Darya [45]

The position of the first ball is

y_1=h-\dfrac g2t^2

while the position of the second ball, thrown with initial velocity v, is

y_2=vt-\dfrac g2t^2

The time it takes for the first ball to reach the halfway point satisfies

\dfrac h2=h-\dfrac g2t^2

\implies\dfrac h2=\dfrac g2t^2

\implies t=\sqrt{\dfrac hg}

We want the second ball to reach the same height at the same time, so that

\dfrac h2=v\sqrt{\dfrac hg}-\dfrac g2\left(\sqrt{\dfrac hg}\right)^2

\implies h=2v\sqrt{\dfrac hg}-g\left(\dfrac hg\right)

\implies h=v\sqrt{\dfrac hg}

\implies v=\sqrt{hg}

8 0
3 years ago
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