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tino4ka555 [31]

4 years ago

12

A car has a momentum of 20,000 kg • m/s. what would the car’s momentum be if its velocity doubles? 10,000 kg • m/s 20,000 kg • m

/s 40,000 kg • m/s 80,000 kg • m/s

2 answers:

babunello [35]4 years ago

8 0

It would definitely be 40,000 <span>kg • m/s because if you double either the mass or velocity and the other stays the same, the momentum will double no matter what.</span>

Jet001 [13]4 years ago

7 0

Answer: 40,000 kg • m/s

Explanation:

the momentum of an object is defined as the product between its mass (m) and its velocity (v):

$p=mv$

From the formula, we see that there is a direct proportionality between the momentum, p, and the velocity, v. Therefore, if the velocity is doubled, then the momentum doubles as well.

In this problem, the initial momentum of the car is 20,000 kg • m/s. Therefore, if the velocity of the car is doubled, the momentum of the car will double as well, and it will becomes

$2\cdot 20,000 kg \cdot m/s=40,000 kg m/s$

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Two wires with the same resistance have the same diameter but different lengths. If wire 1 has length L 1 and wire 2 has length

SVEN [57.7K]

Answer with Explanation:

We are given that

Length of wire 1= $L_1$

Length of wire 2= $L_2$

Resistivity of copper wire= $\rho_1=1.7\times 10^{-5}\Omega-m$

Resistivity of aluminum wire= $\rho_2=2.82\times 10^{-5}\Omega-m$

Wire 1=Copper wire

Wire 2=Aluminum wire

Diameter of both wires are same and resistance of both wires are also same.

We know that

Resistance = $\frac{\rho l}{A}$

When diameter of wires are same then their cross section area are also same .

$l=\frac{RA}{\rho}$

When resistance and area are same then the length of wire depend upon the resistivity of wire .

The length of wire is inversely proportional to resistivity.

When resistivity is greater then the length of wire will be short and when the resistivity is small then the length of wire will be large.

$\rho_1$

Therefore, $L_1>L_2$

Hence, the length of wire 1 (copper wire) is greater than the length of wire 2 (aluminum).

$\frac{L_1}{L_2}=\frac{\frac{RA}{1.7\times 10^{-5}}}{\frac{RA}{2.82\times 10^{-5}}}=1.66$

$L_1=1.66L_2$

7 0

3 years ago

If an X-ray beam of wavelength 1.4 × 10-10 m makes an angle of 30° with a set of planes in a crystal causing first order constru

kkurt [141]

Answer:

Plane spacing, $d=1.4\times 10^{-10}\ m$

Explanation:

It is given that,

Wavelength of x-ray, $\lambda=1.4\times 10^{-10}\ m$

Angle the x-ray made with a set of planes in a crystal causing first order constructive interference is 30 degrees

We need to find the plane spacing. It is based on Bragg's law such that,

$2d\sin\theta=n\lambda$

d is plane spacing

n = 1 here

$d=\dfrac{\lambda}{2\sin\theta}\\\\d=\dfrac{1.4\times 10^{-10}}{2\times \sin (30)}\\\\d=1.4\times 10^{-10}\ m$

So, the plane spacing is $1.4\times 10^{-10}\ m$ .

5 0

3 years ago

A 3" diameter germanium wafer that is 0.020" thick at 300K has 1.015 x 10^17 As atoms added to it. What is the resistivity of th

Ber [7]

Answer:

0.546 ohm / μm

Explanation:

Given that :

N = 1.015 * 10^17

Electron mobility, u = 3900

Hole mobility, h = 1900

Ng = 4.42 x10^22

q = 1.6*10^-19

Resistivity = 1/qNu

Resistivsity (R) = 1/(1.6*10^-19 * 1.015 * 10^17 * 3900)

= 0.01578880889 ohm /cm

Resistivity of germanium :

R = 1 / 2q * sqrt(Ng) * sqrt(u*h)

R = 1 / 2 * 1.6*10^-19 * sqrt(4.42 x10^22) * sqrt(3900*1900)

R = 1 /0.0001831

R = 5461.4964 ohm /cm

5461.4964 / 10000

0.546 ohm / μm

7 0

3 years ago

Name at least three physical properties of the bowling ball.

HACTEHA [7]

The bowling ball is round in shape and red in color.

The bowling ball is smooth and hard.

*The shape of the bowling ball is round.

The color of the bowling ball is red.

The bowling ball is hard.

The bowling ball is smooth.

5 0

3 years ago

Read 2 more answers

A 12cm candle is placed 6cm from a converging lens with a focal length of 15cm. What is the height of the image of the candle? S

amm1812

Answer:

The height of the image of the candle is 20 cm.

Explanation:

Given that,

Size of the candle, h = 12 cm

Object distance from the candle, u = -6 cm

Focal length of converging lens, f = 15 cm

To find,

The height of the image of the candle.

Solution,

Firstly, we will find the image distance of the candle. Let it is equal to v. Using lens formula to find the image distance.

$\dfrac{1}{f}=\dfrac{1}{v}-\dfrac{1}{u}$

v is image distance

$\dfrac{1}{f}=\dfrac{1}{v}-\dfrac{1}{u}\\\\\dfrac{1}{v}=\dfrac{1}{f}+\dfrac{1}{u}\\\\\dfrac{1}{v}=\dfrac{1}{15}+\dfrac{1}{(-6)}\\\\v=-10\ cm$

If h' is the height of the image. Magnification is given by :

$m=\dfrac{h'}{h}=\dfrac{v}{u}$

$h'=\dfrac{vh}{u}\\\\h'=\dfrac{-10\times 12}{-6}\\\\h'=20\ cm$

So, the height of the image of the candle is 20 cm.

3 0

3 years ago