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2 years ago

How long would it take to go from 3m/s to 7m/s when riding a bike if the acceleration was 1.5 m/s/s

Physics

1 answer:

Usimov [2.4K]2 years ago

5 0

Answer: 3 seconds

Explanation: since 3 is 4 away from 7, 3 times 1.5 is 4.5 which is the closest you can get to 7 without being below 7

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A child is riding in a child-restraint chair, securely fastened to the seat of a car. Assume the car has speed 47 km/h when it h

Elina [12.6K]

Answer: F = 1235 N

Explanation: Newton's Second Law of Motion describes the effect of mass and net force upon acceleration: $F_{net}=m.a$

Acceleration is the change of velocity in a period of time: $a=\frac{\Delta v}{\Delta t}$

Velocity of the car is in km/h. Transforming it in m/s:

$v=\frac{47.10^{3}}{36.10^{2}}$

v = 13 m/s

At the moment the car decelerates, acceleration is

$a=\frac{13}{0.2}$

a = 65 m/s²

Then, force will be

$F_{net}=19(65)$

$F_{net}$ = 1235 N

The horizontal net force the straps of the restraint chair exerted on the child to hold her is 1235 newtons.

5 0

2 years ago

Metal sphere A has a charge of − Q . −Q. An identical metal sphere B has a charge of + 2 Q . +2Q. The magnitude of the electric

Goshia [24]

Complete Question:

Metal sphere A has a charge of − Q . −Q. An identical metal sphere B has a charge of + 2 Q . +2Q. The magnitude of the electric force on sphere B due to sphere A is F . F. The magnitude of the electric force on sphere A due to sphere B must be:

A. 2F

B. F/4

C. F/2

D. F

E. 4F

Answer:

Explanation:

If both spheres can be treated as point charges, they must obey the Coulomb's law, that can be written as follows (in magnitude):

$F =\frac{kQ*2Q}{r^{2} }$

As it can be seen, this force is proportional to the product of the charges, so it must be the same for both charges.

As this force obeys also the Newton's 3rd Law, we conclude that the magnitude of the electric force on sphere A due to sphere B, must be equal to the the magnitude of the force on the sphere B due to the sphere A, i.e., just F.

3 0

3 years ago

A ball has a mass of 0.046kg. Calculate the change in gravitational potential energy when the ball is lifted through a vertical

loris [4]

Answer:

PE=0.92414J and KE=0.28175J

Explanation:

Gravitational potential energy=mass*gravity*height

PE=mgh

Data,

M=0.046kg

H=2.05m

g=9.8m/s^2

PE=0.046kg * 9.8m/s^2 * 2.05m

PE =0.92414J

KE=1/2mv^2

M=0.046kg

V=3.5m/s

KE=[(0.046kg)*(3.5m/s)^2]\2

KE=0.28175J

3 0

2 years ago

A box is sliding down an incline tilted at a 12° angle above horizontal. The box is initially sliding down the incline at a spee

OLga [1]

Answer:

The box will cover a distance of 0.9199m before coming to rest

Explanation:

We are given;

Angle of tilt; θ = 12°

Speed of sliding down; u = 1.5 m/s

Coefficient of kinetic friction; μ = 0.34

We are told that the box is sliding down an incline tilted at a 12° angle above horizontal.

Thus,

The components of the weight of the block would be;

Fx = mg sinθ = mg sin 12

Fy = mg cosθ = mg cos 12

For, the normal force on the block, it will be counter balanced by the Y component of weight of block and so we have;

Normal force; Fn = mg cos 12

Now formula for the frictional force would be given by;

Ff = μmg cos 12

So, Ff = 0.34mg cos 12

So, the net force along the inclined plane is;

Fnet = Fx - Ff

Fnet = mg sin 12 - 0.34mg cos 12

Where Fnet = mass x acceleration.

Thus;

ma = mg sin 12 - 0.34mg cos 12

m will cancel out to give;

a = g sin 12 - 0.34g cos 12

a = 9.81(0.2079) - 0.34(9.81 × 0.9781)

a = -1.223 m/s²

According to Newton's equation of motion, we have;

(v² - u²) = 2as

s = (v² - u²)/2a

Final velocity is zero. Thus;

s = (0² - 1.5²)/(2 × -1.223)

s = -2.25/-2.446

s = 0.9199 m

Thus, the box will cover 0.9199m before coming to rest

4 0

3 years ago

The formation of lactic acid in human muscle most closely associated with

zavuch27 [327]

The formation of lactic acid in human muscles is closely associated with intense exertion or activity, during which aerobic respiration that uses oxygen to produce energy cannot be able to match the demand of energy by the muscles. The muscles therefore resort to anaerobic respiration for energy where pyruvate becomes a makeshift hydrogen acceptor rather than oxygen as happens in aerobic respiration. Pyruvate accepts a hydrogen from NADH and becomes reduced to lactate or lactic acid while NADH is oxidized to NAD which is crucial in the formation of energy that is then stored in the form of ATP which is used to re-fuel the muscles to keep them going.

8 0

2 years ago