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2 years ago

Which of the following actions is an example of an improvement in human capital?

Physics

1 answer:

kap26 [50]2 years ago

8 0

Answer: D

Explanation:

Frank’s Furniture trains all of its employees to use computers is an example of an improvement in human capital.

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When a vehicle travels at a steady speed the driving force and the

Ivanshal [37]

Answer:

.......

Must be equal

Explanation:

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2 years ago

A 60kg bicyclist (including the bicycle) is pedaling to the

Fittoniya [83]

a) 4 forces

b) 186 N

c) 246 N

Explanation:

Let's count the forces acting on the bicylist:

1) Weight ( $W=mg$ ): this is the gravitational force exerted on the bicyclist by the Earth, which pulls the bicyclist towards the Earth's centre; so, this force acts downward (m = mass of the bicyclist, g = acceleration due to gravity)

2) Normal reaction (N): this is the reaction force exerted by the road on the bicyclist. This force acts vertically upward, and it balances the weight, so its magnitude is equal to the weight of the bicyclist, and its direction is opposite

3) Applied force ( $F_A$ ): this is the force exerted by the bicylicist to push the bike forward. Its direction is forward

4) Air drag ( $R$ ): this is the force exerted by the air on the bicyclist and resisting the motion of the bike; its direction is opposite to the motion of the bike, so it is in the backward direction

So, we have 4 forces in total.

Here we can find the net force on the bicyclist by using Newton's second law of motion, which states that the net force acting on a body is equal to the product between the mass of the body and its acceleration:

$F_{net}=ma$

where

$F_{net}$ is the net force

m is the mass of the body

a is its acceleration

In this problem we have:

m = 60 kg is the mass of the bicyclist

$a=3.1 m/s^2$ is its acceleration

Substituting, we find the net force on the bicyclist:

$F_{net}=(60)(3.1)=186 N$

We can write the net force acting on the bicyclist in the horizontal direction as the resultant of the two forces acting along this direction, so:

$F_{net}=F_a-R$

where:

$F_{net}$ is the net force

$F_a$ is the applied force (forward)

$R$ is the air drag (backward)

In this problem we have:

$F_{net}=186 N$ is the net force (found in part b)

$R=60 N$ is the magnitude of the air drag

Solving for $F_a$ , we find the force produced by the bicyclist while pedaling:

$F_a=F_{net}+R=186+60=246 N$

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3 years ago

At what speed do a bicycle and its rider, with a combined mass of 100 kg , have the same momentum as a 1600 kg car traveling at

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$Given:\\m_b=100kg\\m_c=1600kg\\v_c=5.2 \frac{m}{s} \\p_b=p_c\\\\Find:\\v_b=?\\\\Solution:\\\\p_b=p_c\\\\p=mv\\\\m_bv_b=m_cv_c\Rightarrow v_b= \frac{m_cv_c}{m_b} \\\\v_b= \frac{1600kg\cdot5.2 \frac{m}{s} }{100kg} =83.2 \frac{m}{s}$

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3 years ago

Pick an activity you enjoy, such as running or riding a scooter, and describe how newton's laws apply to that activity.

deff fn [24]

I like playing basketball. So I'm the object in motion. Until an unbalanced force comes and hits me I fall and stay at rest.

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3 years ago

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A boy is running to the east and jumps into a stationary boat in a lake. Once the boy lands in the boat, the boat moves to the e

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<span>The initial momentum of the boy is equal to the boy/boat because the final velocity of the boat is less than the initial velocity of the boy.</span>

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3 years ago