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2 years ago

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d. The force is doubled and the object’s mass is halved? 18. ||| A man pulling an empty wagon causes it to accelerate at 1.4 m/s

2. What will the acceleration be if he pulls with the same force when the wagon contains a child whose mass is three times that of the wagon?

1 answer:

Harrizon [31]2 years ago

8 0

Answer:

$a' = 0.35 m/s^2$

Explanation:

Let say the empty wagon has mass "M"

now by newton's II Law we will have

$F = Ma$

now it is given that empty wagon is pulled with acceleration 1.4 m/s/s

now we will have

$F = 1.4 M$

now a child of mass three times the mass of wagon is sitting on the empty wagon

so here we have

$F = (M + 3M) a$

$1.4 M = 4M a'$

so we have

$a' = 0.35 m/s^2$

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If the newton is the product of kilograms and meters/second2 what units comprise the pound?

Kobotan [32]

Answer:

Pound is the product of slug and foot/square second.

Explanation:

We are given that

Force=1 N

1N= $1kg\times ms^{-2}$

We have to find the units comprise the pound.

Force=1 Pound

Mass=Slug

Acceleration= $ft/s^2$

Therefore,

1 pound= $1 slug\times fts^{-2}$

Therefore, we can write as 1 pound is equal to the product of slug and ft/square second.

Hence, pound is the product of slug and foot/square second.

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2 years ago

How much power does a machine have that does 15204 J of work in 40 seconds?

pentagon [3]

Explanation:

since P=W/t

P=15204/40

P=380.1 Watt

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1 year ago

A stretched spring has 5184 J of elastic potential energy and a spring constant of 16,200 N/m. What is the displacement of the s

Bezzdna [24]

Hello!

A stretched spring has 5184 J of elastic potential energy and a spring constant of 16,200 N/m. What is the displacement of the spring?

0.57 m

0.64 m

0.80 m

1.25 m

Data:

$E_{pe}\:(elastic\:potential\:energy) = 5184\:J$

$K\:(constant) = 16200\:N/m$

$x\:(displacement) =\:?$

For a spring (or an elastic), the elastic potential energy is calculated by the following expression:

$E_{pe} = \dfrac{k*x^2}{2}$

Where k represents the elastic constant of the spring (or elastic) and x the deformation or displacement suffered by the spring.

Solving:

$E_{pe} = \dfrac{k*x^2}{2}$

$5184 = \dfrac{16200*x^2}{2}$

$5184*2 = 16200*x^2$

$10368 = 16200\:x^2$

$16200\:x^2 = 10368$

$x^{2} = \dfrac{10368}{16200}$

$x^{2} = 0.64$

$x = \sqrt{0.64}$

$\boxed{\boxed{x = 0.8\:m}}\end{array}}\qquad\checkmark$

Answer:

The displacement of the spring = 0.8 m (or 0.80 m)

_________________________________________

I Hope this helps, greetings ... Dexteright02! =)

8 0

2 years ago

Read 2 more answers

The moon’s orbital speed around Earth is 3.680 × 10^3 km/h. Suppose the moon suffers a perfectly elastic collision with a comet

Naily [24]

Answer:

Speed of comet before collision is

$v_{2_{i}}=-2.5\times10^{3}\quad km/h$

Explanation:

Correction: (As stated after collision comet moves away from moon so velocity of moon and moon and comet must be opposite in direction. as spped of moon after collision is −4.40 × 10^2km/h so that comet's must be 5.740 × 10^3km/h instead of -5.740 × 10^3km/h)

Solution:

$mass \quad of\quad moon = m_{1}\\\\mass\quad of \quad comet = m_{2} = 0.5 m_{1}\\\\speed\quad of\quad moon\quad before\quad collision = v_{1_{i}}=3.680\times 10^3 km/h\\\\speed \quad of\quad moon\quad after\quad collision=v_{1_{f}} = -4.40 \times 10^2 km/h\\\\speed\quad of\quad comet\quad after\quad collision =v_{2_{f}} =5.740 \times 10^3 km/h$

Case is considered as partially inelastic collision, by conservation of momentum

$m_{1}v_{1_{i}}+m_{2}v_{2_{i}}=m_{1}v_{1_{f}}+m_{2}v_{2_{f}}\\\\m_{1}v_{1_{i}}+0.5m_{1}v_{2_{i}}=m_{1}v_{1_{f}}+0.5m_{1}v_{2_{f}}\\\\v_{1_{i}}+0.5v_{2_{i}}=v_{1_{f}}+0.5v_{2_{f}}\\\\v_{2_{i}}=2(v_{1_{f}}+0.5v_{2_{f}}-v_{1_{i}})\\\\v_{2_{i}}=2(-4.40 \times 10^2+0.5(5.740 \times 10^3)-3.680 \times 10^3 )\\\\v_{2_{i}}=-2.5\times10^{3}\quad km/h$

8 0

2 years ago

In an experiment to measure the wavelength of light using a double slit, it is found that the fringes are too close together to

Mandarinka [93]

Answer:

halve the slit separation

Explanation:

As we know that

In YDS experiment, the equation of fringe width is as follows

$\beta = \frac{\lambda D}{d}$

where,

D denotes the separation in the middle of screen and slits

d denotes the distance in the middle of two slits

And to increase the Δx we have to decrease the d i.e, the distance between the two slits

Hence, the first option is correct

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2 years ago