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3 years ago

What will the motion of the cart be like when there is no force at all ? ( There is no friction in this model. )

Physics

1 answer:

Nezavi [6.7K]3 years ago

7 0

Answer:

Explanation:

the cart motion will be backward motion for there is no action force generated via friction force which acts on the ground harness of the horse. thus the will be pulled backward by the cart.

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2.<br> ___ ___ is the number of wavelengths that pass a given point/second.

atroni [7]

Answer:

frequency tell me if im right

Explanation:

5 0

3 years ago

PLZ help this should be super easy

Ugo [173]

The answer is number (3) if am wrong am sry

4 0

3 years ago

Read 2 more answers

How fast would a 2-kilogram object need to move to have the same kinetic

siniylev [52]

Answer:

11.3 m/s

Explanation:

KE₁ = KE₂

½m₁v₁² = ½m₂v₂²

½ (2 kg) v² = ½ (4 kg) (8 m/s)²

v ≈ 11.3 m/s

8 0

3 years ago

What amount of heat is required to raise the temperature of 200 g of water by 15°C (the specific heat of water is 1 cal/g°C)

Sergeeva-Olga [200]

Answer:

Heat energy required (Q) = 3,000 J

Explanation:

Find:

Mass of water (M) = 200 g

Change in temperature (ΔT) = 15°C

Specific heat of water (C) = 1 cal/g°C

Find:

Heat energy required (Q) = ?

Computation:

Q = M × ΔT × C

Heat energy required (Q) = Mass of water (M) × Change in temperature (ΔT) × Specific heat of water (C)

Heat energy required (Q) = 200 g × 15°C × 1 cal/g°C

Heat energy required (Q) = 3,000 J

4 0

3 years ago

The opening to a cave is a tall, 30.0-cm-wide crack. A bat is preparing to leave the cave emits a 30.0 kHz ultrasonic chirp. How

Vlada [557]

Answer:

The value is $w = 7.54 \ m$

Explanation:

From the question we are told that

The length of the crack is $a = 0.3 \ m$

The frequency is $f = 30.0 \ kHz = 30 *10^{3} \ Hz$

The distance outside the cave that is being consider is $D = 100 \ m$

The speed of sound is $v_s = 340 \ m/s$

Generally the wavelength of the wave is mathematically represented as

$\lambda = \frac{v}f}$

=> $\lambda = \frac{340 }{30*10^{3}}$

=> $\lambda = 0.0113 \ m/s$

Generally for a single slit the path difference between the interference patterns of the sound wave and the center is mathematically represented as

$y = \frac{ n * \lambda * D}{a}$

=> $y = \frac{ 1 * 0.0113 * 100}{0.3}$

=> $y = 3.77 \ m$

Generally the width of the sound beam is mathematically represented as

$w = 2 * y$

=> $w = 2 * 3.77$

=> $w = 7.54 \ m$

4 0

3 years ago