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3 years ago

A. A horse pulls a cart along a flat road. Consider the following four forces that arise in this situation.

Physics

1 answer:

never [62]3 years ago

3 0

Answer:

a) F₁ = F₂, F₃ = F₄, b) the correct answer is 3

Explanation:

a) In this exercise we have several action and reaction forces, which are characterized by having the same magnitude, but different direction and being applied to different bodies

Forces 1 and 2 are action and reaction forces F₁ = F₂

Forces 3 and 4 are action and reaction forces F₃ = F₄

as it indicates that the

b) how the car increases if speed implies that force 1> force3

F₁ > F₃

therefore the correct answer is 3

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A sled that has a mass of 8 kg is pulled at a 50 degree angle with a force of 20 N. The force of friction acting on the sled is

vitfil [10]

The acceleration of the sled will be 1.30 m/s². Force is defined as the product of mass and acceleration.

<h3>What is force?</h3>

Force is defined as the push or pulls applied to the body. Sometimes it is used to change the shape, size, and direction of the body.

Given data;

m(mass of sled)=8 kg

Θ is the inclination of force= 50°

Force of friction,f=2.4 N.

The applied force at the given angle is resolved into the two-component as;

$\rm F_h=F cos \theta \\\\ F_h= 20 cos 50 ^0 \\\\ F_h= 12.85 \ N$

$\rm F_v=F sin \theta \\\\ F_v=20 sin 50^0 \\\\ F_v=15.32 \ N$

The net vertical force is zero;

$\rm F_N=mg-Fsin50^0 \\\\ \rm F_N=8 \times 9.81 -15.32 \\\\ F_N=63.1 \ N \\\\$

From Newton's second law the net force as;

$\rm \sum F_{net}=ma \\\\ Fcos 60^0-f =ma \\\\ a=\frac{12.855-2.4}{8} \\\\a = 1.30 \ m/s^2$

Hence, the acceleration of the sled will be 1.30 m/s².

To learn more about the force refer to the link;

brainly.com/question/26115859

#SPJ1

5 0

2 years ago

Read 2 more answers

A bat moving at 3.7 m/s is chasing a ying insect. The bat emits a 36 kHz chirp and receives back an echo at 36.79 kHz. At what s

3241004551 [841]

Answer:

The speed the bat is gaining on its prey is 0.03m/s

Explanation:

Given;

speed of the bat, v₀ = 3.7 m/s

frequency of the bat, F₀ = 36 kHz

frequency of the source, Fs = 36.79

This is relative motion between a source of the sound and the observer. The phenomenon is known as Doppler effect.

Apply the following equation to determine the speed of the insect which is the source;

$F_0 = F_s[\frac{v+v_0}{v-v_s} ]\\\\\frac{F_0}{F_s} = [\frac{v+v_0}{v-v_s} ]\\\\\frac{36.79}{36} = \frac{340+3.7}{340-v_s}\\\\1.0219 = \frac{343.7}{340-v_s}\\\\ 340-v_s = \frac{343.7}{1.0219}\\\\340-v_s = 336.33\\\\v_s = 340-336.33\\\\v_s = 3.67 \ m/s$