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kirill [66]
3 years ago
7

A truck has shock absorbers with a spring constant of 24200 N/m. When it hits a bump, it oscillates at 0.429 Hz. What is the mas

s of the truck? (Unit=kg)
Physics
1 answer:
siniylev [52]3 years ago
3 0

Answer:

3331.5 kg

Explanation:

Given:

Spring constant of the spring (k) = 24200 N/m

Frequency of oscillation (f) = 0.429 Hz

Let the mass be 'm' kg.

Now, we know that, a spring-mass system undergoes Simple Harmonic Motion (SHM). The frequency of oscillation of SHM is given as:

f=\frac{1}{2\pi}\sqrt{\frac{k}{m}}

Rewrite the above equation in terms of 'm'. This gives,

2\pi f=\sqrt{\frac{k}{m}}\\\\Squaring\ both\ sides,\ we\ get:\\\\(2\pi f)^2=\frac{k}{m}\\\\m=\frac{k}{4\pi^2 f^2}

Now, plug in the given values and solve for 'm'. This gives,

m=\frac{24200\ N/m}{4\pi^2\times (0.429\ Hz)^2 }\\\\m=\frac{24200\ N/m}{4\pi^2\times 0.184\ Hz^2}\\\\m\approx3331.5\ kg

Therefore, the mass of the truck is 3331.5 kg.

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Answer:

0.15 s

Explanation:

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

Speed of sound (v) = 330 m/s

Distance (x) = 25 m

Time (t) =?

The time taken for the echo of the sound to the bat can be obtained as follow:

v = 2x / t

330 = 2 × 25 / t

330 = 50 / t

Cross multiply

330 × t = 50

Divide both side by 330

t = 50 / 330

t = 0.15 s

Thus, it will take 0.15 s for the echo of the sound to the bat

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3 years ago
An empty truck traveling at 10 km/h has kinetic energy. How much kinetic energy does it have when loaded so that its mass and it
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Answer:

8 time increase in K.E.

Explanation:

Consider Mass of truck = m kg and speed = v m/s then

K.E. = 1/2 ×mv²

If mass and speed both are doubled i.e let m₀ = 2m and v₀ = 2v then

(K.E.)₀ =  1/2 ×2m(2v)²

(K.E.)₀ = 8 (1/2 × mv²) = 8 × K.E.

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3 years ago
Suppose 8.50 ✕ 10^5 J of energy are transferred to 1.63 kg of ice at 0°C. The latent heat of fusion and specific heat of water a
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Answer:

(a) 5.43 x 10⁵ J

(b) 3.07 x 10⁵ J

(c) 45 °C

Explanation:

(a)

L_{f} = Latent heat of fusion of ice to water = 3.33 x 10⁵ J/kg

m = mass of ice = 1.63 kg

Q_{f} = Energy required to melt the ice

Energy required to melt the ice is given as

Q_{f} = m L_{f}

Q_{f} = (1.63) (3.33 x 10⁵)

Q_{f} = 5.43 x 10⁵ J

(b)

E = Total energy transferred = 8.50 x 10⁵ J

Q  = Amount of energy remaining to raise the temperature

Using conservation of energy

E = Q_{f} + Q

8.50 x 10⁵ = 5.43 x 10⁵ + Q

Q = 3.07 x 10⁵ J

(c)

T₀ = initial temperature = 0°C

T = Final temperature

m = mass of water = 1.63 kg

c = specific heat of water = 4186 J/(kg °C)

Q = Amount of energy to raise the temperature of water = 3.07 x 10⁵ J

Using the equation

Q = m c (T - T₀)

3.07 x 10⁵ = (1.63) (4186) (T - 0)

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A dentist’s drill starts from rest. After 3.20 s of constant angu-lar acceleration, it turns at a rate of 2.51 3 104 re v/m i n.
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Answer:

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w_{i} means initial angular velocity, which is 0 rev/min

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w_{f}= 2628.5 rad/s

so the angular acceleration, α will be:

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a = 821.40 rad/s^{2}

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ΔTita = \frac{w^{2} _{f}-w^{2} _{i}  }{2a}

ΔTita = ((2628.5)^{2} - (0 rev/min)^{2}) / 2* 821.40

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