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charle [14.2K]
2 years ago
8

Which two statements describe force?

Physics
1 answer:
Len [333]2 years ago
6 0

Answer:

I think its C and D

Explanation:Force is like a energy that moves stuff. :)

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A city planner is working on the redesign of a hilly portion of a city. An important consideration is how steep the roads can be
Artyom0805 [142]

Explanation:

It is known that relation between force and acceleration is as follows.

                      F = m \times a

I is given that, mass is 1090 kg and acceleration is 21 m/s. Therefore, we will calculate force as follows.

              F = m \times a      

                 = 1090 \times (\frac{21}{16})

                 = 1430.625 N

Also, it is known that

      sin(\theta) = \frac{\text{Force car can exert}}{\text{Force gravity pulls car}}

      sin(\theta) = \frac{1430.625 N}{(1090 \times 9.8) N}

        \theta = 7.70 degrees

Thus, we can conclude that the maximum steepness for the car to still be able to accelerate is 7.70 degrees.

8 0
3 years ago
A 0.0450 kg bullet is accelerated from rest to a speed of 425 m/s in a 2.25 kg rifle (which is inititally at rest). The pain of
Mrac [35]

Answer:

If the rifle is held loosely away from the shoulder, the recoil velocity will be of -8.5 m/s, and the kinetic energy the rifle gains will be 81.28 J.

Explanation:

By momentum conservation, <em>and given the bullit and the recoil are in a straight line</em>, the momentum analysis will be <em>unidimentional</em>. As the initial momentum is equal to zero (the masses are at rest), we have that the final momentum equals zero, so

0=P_{f}=m_{b} *v_{b}+m_{r}*v_{r}

now we clear v_{r} and use the given data to get that

v_{r}=-8.5\frac{m}{s}

<em>But we have to keep in mind that the bullit accelerate from rest to a speed of 425 m/s</em>, then <u>if the rifle were against the shoulder, the recoil velocity would be a fraction of the result obtained</u>, but, as the gun is a few centimeters away from the shoulder, it is assumed that the bullit get to its final velocity, so the kick of the gun, gets to its final velocity \bold{v_{r}} too.

Finally, using v_{r} we calculate the kinetic energy as

K=\frac{1}{2}m_{r}v_{r}^{2}=81.28J

3 0
3 years ago
The work-energy principle states that the work done by all the __________ forces acting on an object (or system of objects) caus
CaHeK987 [17]

The work-energy principle states that the work done by all the non-conservative forces acting on an object (or system of objects) causes a change in the total mechanical energy of the object or system.


What is the work-energy principle?
The work-energy principle states that the total work done on a system is equal to the change in kinetic energy of the system. It is given as:

W.D = ΔK.E

       = K.E₁ - K.E₂

where K.E₁ is the initial kinetic energy of the system

            K.E₂ is the final kinetic energy of the system



What is meant by non-conservative forces?

Non-conservative forces as the name suggests are not conserved i.e. these forces cause a loss of mechanical energy from the system. A prime example of non-conservative forces is friction.

The total mechanical energy of the system is the sum of the potential energy and kinetic energy that the system contains. This energy is conserved and follows the work-energy theorem.

Learn more about work and energy here:

<u>brainly.com/question/17290830</u>

#SPJ4

4 0
2 years ago
The asteroid, Ida, has a small moon, Dactyl, that orbits at a speed of 5.66 m/s in an orbit of radius 90, 000m . What is Ida's m
cricket20 [7]

Answer:

4.3 x 10^16 kg

Explanation:

M = rv^2/G =[90,000 x 5.66^2] / [6.67 x 10^-11]

M = 43,226,446,776,611,694 = 4.3 x 10^16 kg - Ida's mass.

4 0
3 years ago
n the figure, a uniform ladder 12 meters long rests against a vertical frictionless wall. The ladder weighs 400 N and makes an a
Leto [7]

Since the ladder is standing, we know that the coefficient of friction is at least something. This [gotta be at least this] friction coefficient can be calculated. As the man begins to climb the ladder, the friction can even be less than the free-standing friction coefficient. However, as the man climbs the ladder, more and more friction is required. Since he eventually slips, we know that friction is less than what's required at the top of the ladder.

 

The only "answer" to this problem is putting lower and upper bounds on the coefficient. For the lower one, find how much friction the ladder needs to stand by itself. For the most that friction could be, find what friction is when the man reaches the top of the ladder.

Ff = uN1

Fx = 0 = Ff + N2

Fy = 0 = N1 – 400 – 864

N1 = 1264 N

Torque balance

T = 0 = N2(12)sin(60) – 400(6)cos(60) – 864(7.8)cos(60)

N2 = 439 N

Ff = 439= u N1

U = 440 / 1264 = 0.3481

3 0
3 years ago
Read 2 more answers
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