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3 years ago

Newton’s second law of motion addresses the relationship between what two variables that influence the force on a body?

Physics

2 answers:

Eddi Din [679]3 years ago

5 0

Answer: a

Explanation:

Helen [10]3 years ago

3 0

Answer:

A. Mass and acceleration

Explanation:

According to Newton's second law of motion, the resultant force is directly proportional to the rate of change in momentum
Therefore; F = ma , where F is the resultant force, m is the mass, and a is the acceleration of the body.
<u>Resultant force depends on the acceleration and the mass of a body in motion, an increase in acceleration causes a corresponding increase in resultant force.</u> A body with higher mass will have a larger strong force if the acceleration is kept constant.

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An open system starts with 52 J of mechanical energy. The energy changes

spayn [35]

Answer:

I think the answer is D,54 joules

5 0

3 years ago

you have been called to testify as a as an expert witness in a trial involving a head-on collision Car A weighs 1515 pounds and

GrogVix [38]

Answer:

70.6 mph

Explanation:

Car A mass= 1515 lb

Car B mass=1125 lb

Speed of car B is 46 miles/h

Distance before locking, d=19.5 ft

Coefficient of kinetic friction is 0.75

Initial momentum of car B=mv where m is mass and v is velocity in ft/s

46 mph*1.46667=67.4666668 ft/s

$Momentum_B=1125*67.4666668 ft/s$

Initial momentum of car A is given by

$Momentum_A=1515v_a$ where $v_a$ is velocity of A

Taking East as positive and west as negative then the sum of initial momentum is

$1515v_a-(1125*67.4666668 ft/s)$

The common velocity is represented as $v_c$ hence after collision, the final momentum is

$Momentum_final=(m_a+m_b)v_c=(1515+1125)v_c=2640v_c$

From the law of conservation of linear momentum, sum of initial and final momentum equals each other hence

$1515v_a-(1125*67.4666668 ft/s)= 2640v_c$

The acceleration of two cars $a=-\mug=-0.75*32.17=-24.1275 ft/s^{2}$

From kinematic equation

$v^{2}=u^{2}+2as$ hence

$v^{2}-u^{2}=2as$

$0^{2}-(v_c)^{2}=2*-24.1275*19.5$

$v_c=\sqrt{2*24.1275*19.5}=30.67 ft/s$

Substituting the value of $v_c$ in equation $1515v_a-(1125*67.4666668 ft/s)= 2640v_c$

$1515v_a-(1125*67.4666668 ft/s)= 2640*30.67$

$1515v_a=(1125*67.4666668 ft/s)+2640*30.67$

$v_a=\frac {156868.8}{1515}=103.5438 ft/s$

$\frac {103.5438}{1.46667}=70.59787 mph\approx 70.60 mph$

3 0

3 years ago

You are trying to overhear a juicy conversation, but from your distance of 23.0 m , it sounds like only an average whisper of 40

Len [333]

If its asking the distance for the 65 db then use a proportion, if otherwise pleas clarify. It sounds like a pretty juicy conversation.

4 0

3 years ago

A spring scale hung from the ceiling stretches by 5.2 cm when a 2.0 kg mass is hung from it. The 2.0 kg mass is removed and repl

pantera1 [17]

Physics - Damon, Wednesday, December 9, 2015 at 5:13am
F = k x

k = 2 g/6.1 cm

2.5g = (2g/6.1cm) x

x = 6.1 (2.5/2) cm

8 0

3 years ago

Read 2 more answers

A 38.2 kg wagon is towed up a hill inclined at 17.5 ◦ with respect to the horizontal. The tow rope is parallel to the incline an

Tema [17]

Answer:

v = 8.57 m/s

Explanation:

As we know that the wagon is pulled up by string system

So the net force on the wagon along the inclined is due to tension in the rope and component of weight along the inclined plane

So as per work energy theorem we know that

work done by tension force + work done by force of gravity = change in kinetic energy

$F_t . d - (mgsin\theta)(d) = \frac{1}{2}mv^2 - 0$

so we have

$F_t = 129 N$

$\theta = 17.5^o$

m = 38.2 kg

d = 85.4 m

so now we have

$129(85.4) - (38.2)9.8sin17.5 (85.4) = \frac{1}{2}(38.2) v^2$

$v = 8.57 m/s$

7 0

3 years ago