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3 years ago

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A 40 kg bike traveling down the road has a negative acceleration. What is true of the force? A. It is positive B. It doesn't exi

st C. It is negative D. It is zero
Help me pls.

2 answers:

Greeley [361]3 years ago

4 0

Answer:

C. It is negative

Explanation:

Per Newton's second law, the net force is the mass times the acceleration:

∑F = ma

If the acceleration is negative, the net force is negative.

S_A_V [24]3 years ago

3 0

Answer:

C. It is negative

Explanation:

As we know that mass of the bike is given as

m = 40 kg

now we know by Newton's law of motion

$F = ma$

here we know that

m = mass of the bike

a = acceleration of the bike

so here we will have

$F = 40 (a )$

now if the acceleration of the bike is in negative direction

then we will get a negative force on it as per above formula

so correct answer will be

C. It is negative

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A ball is thrown up into the air with an initial velocity of 18 m/s. A) How high does the ball go? B) Calculate the time needed

kaheart [24]

Answer:

B) t = 1.83 [s]

A) y = 16.51 [m]

Explanation:

To solve this problem we must use the following equation of kinematics.

$v_{f} =v_{o} -g*t$

where:

Vf = final velocity = 0

Vo = initial velocity = 18 [m/s]

g = gravity acceleration = 9.81 [m/s²]

t = time [s]

Note: the negative sign in the above equation means that the acceleration of gravity is acting in the opposite direction to the motion.

A) The maximum height is reached when the final velocity of the ball is zero.

0 = 18 - (9.81*t)

9.81*t = 18

t = 18/9.81

t = 1.83 [s], we found the answer for B.

Now using the following equation.

$y = y_{o} + v_{o}*t - 0.5*g*t^{2}\\$

where:

y = elevation [m]

Yo = initial elevation = 0

y = 18*(1.83) - 0.5*9.81*(1.83)²

y = 16.51 [m]

7 0

3 years ago

Two particles, each of mass m, are initially at rest very far apart.Obtain an expression for their relative speed of approach at

PSYCHO15rus [73]

Answer:

$|\Delta v |=\sqrt{\frac{4Gm}{d} }$

Explanation:

Consider two particles are initially at rest.

Therefore,

the kinetic energy of the particles is zero.

That initial K.E. = 0

The relative velocity with which both the particles are approaching each other is Δv and their reduced masses are

$\mu= \frac{m_1m_2}{m_1+m_2}$

now, since both the masses have mass m

therefore,

$\mu= \frac{m^2}{2m}$

= m/2

The final K.E. of the particles is

$KE_{final}=\frac{1}{2}\times \mu\times \Delta v^2$

Distance between two particles is d and the gravitational potential energy between them is given by

$PE_{Gravitational}= \frac{Gmm}{d}$

By law of conservation of energy we have

$KE_{initial}+KE_{final}= PE_{gravitaional}$

Now plugging the values we get

$0+\frac{1}{2}\frac{m}{2}\Delta v^2= -\frac{Gmm}{d}$

$|\Delta v |=\sqrt{\frac{4Gm}{d} }$

$=\sqrt{\frac{Gm}{d} }$

This the required relation between G,m and d

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3 years ago

A one-piece cylinder has a core section protruding from the larger drum and is free to rotate around its central axis. A rope wr

PilotLPTM [1.2K]

Answer:

Magnitude the net torque about its axis of rotation is 2.41 Nm

Solution:

As per the question:

The radius of the wrapped rope around the drum, r = 1.33 m

Force applied to the right side of the drum, F = 4.35 N

The radius of the rope wrapped around the core, r' = 0.51 m

Force on the cylinder in the downward direction, F' = 6.62 N

Now, the magnitude of the net torque is given by:

$\tau_{net} = \tau + \tau'$

where

$\tau$ = Torque due to Force, F

$\tau'$ = Torque due to Force, F'

$tau = F\times r$

$tau' = F'\times r'$

Now,

$\tau_{net} = - F\times r + F'\times r'$

$\tau_{net} = - 4.35\times 1.33 + 6.62\times 0.51 = - 2.41\ Nm$

The net torque comes out to be negative, this shows that rotation of cylinder is in the clockwise direction from its stationary position.

Now, the magnitude of the net torque:

$|\tau_{net}| = 2.41\ Nm$

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3 years ago

What are some INDIRECT applications of centripetal force in real life?

Eddi Din [679]

Answer:

The circular turning of roads

Explanation:

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8 0

2 years ago

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Rex decides to launch a priceless vase into the air at a 90-degree angle. The initial velocity vase is +7.0 m/s. Dylan claims th

Sergeu [11.5K]

Answer:

D. Dylan is incorrect because a 90-degree launch angle results in the largest vertical range

Explanation:

Projectile is the motion of an object thrown into space. When an object is thrown into space, the only force which acts on it is the acceleration due to gravity.

An object thrown into space would reach maximum height (vertical range) if it is launched at an angle of 90 degrees. For maximum horizontal range, the object needs to be launched at an angle of 45 degrees.

Therefore Dylan is incorrect because a 90-degree launch angle results in the largest vertical range

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3 years ago