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2 years ago

If the loss of 3500 kcal is equal to a loss of 1.0 lb, how many days will it take charles to lose 5.0 lb

1 answer:

4 0

We are missing an important piece of information needed to answer this question: the number of kcal Charles losses per day. However, we can come up with a general equation in which kcal/day is the only independent variable.

We know that it takes 3500 kcal to lose one pound. To lose 5 pounds, Charles needs to lose 5 x 3500 kcal = 17,500 kcal.

To find how many days it takes Charles to lose 17,500 kcal (5 pounds), we must divide that amount by the number of kcal Charles loses per day.
Here is the equation to calculate that number

Number of days= 17500 / (kcal per day)

If given calories, remember that 1000 calories = 1 kcal, and .001 kcal = 1 cal

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A 2-kg cart, traveling on a horizontal air track with a speed of 3 m/s, collides with a stationary 4-kg cart. The carts stick to

daser333 [38]

Answer:

Magnitude of impulse, |J| = 4 kg-m/s

Explanation:

It is given that,

Mass of cart 1, $m_1=2\ kg$

Mass of cart 2, $m_2=4\ kg$

Initial speed of cart 1, $u_1=3\ m/s$

Initial speed of cart 2, $u_2=0$ (stationary)

The carts stick together. It is the case of inelastic collision. Let V is the combined speed of both carts. The momentum remains conserved.

$m_1u_1+m_2u_2=(m_1+m_2)V$

$V=\dfrac{m_1u_1+m_2u_2}{(m_1+m_2)}$

$V=\dfrac{2\times 3}{(2+4)}$

V = 1 m/s

The magnitude of the impulse exerted by one cart on the other is given by:

$J=F\times t=m(V-u)$

$J=m(V-u)$

$J=2\times (1-3)$

J = -4 kg-m/s

|J| = 4 kg-m/s

So, the magnitude of the impulse exerted by one cart on the other 4 kg-m/s. Hence, this is required solution.

8 0

2 years ago

A 60-kg motor sits on four cylindrical rubber blocks. Each cylinder has a height of 3 cm and a cross-sectional area of 15 cm2. T

stealth61 [152]

<span>A.) If a sideways force of 300 N is applied to the motor, how far will it move sideways?</span>

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2 years ago

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A 5.00-g bullet is shot through a 1.00-kg wood block suspended on a string 2.00 m long. The center of mass of the block rises a

o-na [289]

Answer:395.6 m/s

Explanation:

Given

mass of bullet $m=5 gm$

mass of wood block $M=1 kg$

Length of string $L=2 m$

Center of mass rises to an height of $0.38 cm$

initial velocity of bullet $u=450 m/s$

let $v_1$ and $v_2$ be the velocity of bullet and block after collision

Conserving momentum

$mu=mv_1+Mv_2$ -------------1

Now after the collision block rises to an height of 0.38 cm

Conserving Energy for block

kinetic energy of block at bottom=Gain in Potential Energy

$\frac{Mv_2^2}{2}=Mgh_{cm}$

$v_2=\sqrt{2gh_{cm}}$

$v_2=\sqrt{2\times 9.8\times 0.38}$

$v_2=0.272 m/s$

substitute the value of $v_2$ in equation 1

$5\times 450=5\times v_1+1000\times 0.272$

$v_1=395.6 m/s$

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2 years ago

A car increases its speed from 9.6 meters per second to 11.2 meters per second in 4.0 seconds. The average acceleration of the c

Gwar [14]

The acceleration is the change of speed/velocity over time. Thus to calculate this you do (V1-V2)/T or (11.2-9.6)/4 or 0.4 m/s^2

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2 years ago

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3. What conclusion can you make about the electric field strength between two parallel plates? Explain your answer referencing P

KIM [24]

Answer:

From the relation above we can conclude that the as the distance between the two plate increases the electric field strength decreases

Explanation:

I cannot find any attached photo, but we can proceed anyways theoretically.

The electric field strength (E) at any point in an electric field is the force experienced by a unit positive charge (Q) at that point

i.e

$E=\frac{F}{Q}$

But the force F

$F= \frac{kQ1Q2}{r^2}$

But the electric field intensity due to a point charge Q at a distance r meters away is given by

$E= \frac{\frac{kQ1Q2}{r^2}}{Q} \\\\\E= \frac{Q1}{4\pi er^2 }$

<em>From the relation above we can conclude that the as the distance between the two plate increases the electric field strength decreases</em>

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2 years ago