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3 years ago

Similarity of gravity and electromagnetic force

Physics

1 answer:

Mashutka [201]3 years ago

5 0

Answer:

Gravity is an attractive force as well as electromagnetic, but electromagnetic attracts and repels.

Explanation:

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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

babunello [35]

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>

8 0

3 years ago

Read 2 more answers

If the absolute temperature of a gas is 600 K, the temperature in degrees Celsius is: A. 705Â°C. B. 873Â°C. C. 273Â°C. D. 327Â°C

zlopas [31]

Answer:

D). $327 ^0 C$

Explanation:

As we know that temperature scale is linear so we will have

$\frac{^0C - 0}{100 - 0} = \frac{K - 273}{373 - 273}$

now we have

$\frac{^0 C - 0}{100} = \frac{K - 273}{100}$

so the relation between two scales is given as

$^0 C = K - 273$

now we know that in kelvin scale the absolute temperature is 600 K

so now we have

$T = 600 - 273 = 327 ^0 C$

so correct answer is

D). $327 ^0 C$

4 0

2 years ago

According to one set of measurements, the tensile strength of hair is 196 MPa , which produces a maximum strain of 0.380 in the

slega [8]

Answer:

(a). The magnitude of the force is 0.38416 N.

(b). The original length is 0.0869 m.

Explanation:

Given that,

Tensile strength = 196 MPa

Maximum strain = 0.380

Diameter = 50.0 μm

Length = 12.0 cm

We need to calculate the area

Using formula of area

$A=\dfrac{\pi}{4}\times d^2$

Put the value into the formula

$A=\dfrac{\pi}{4}\times(50.0\times10^{-6})^2$

$A=1.96\times10^{-9}\ m^2$

We need to calculate the magnitude of the force

Using formula of force

$F=\sigma A$

Put the value into the formula

$F=196\times10^{6}\times1.96\times10^{-9}$

$F=0.38416\ N$

(b). If the length of a strand of the hair is 12.0 cm at its breaking point

We need to calculate the unstressed length

Using formula of strain

$strain=\dfrac{\Delta l}{l_{0}}$

$\Delta l=strain\times l_{0}$

Put the value into the formula

$\Delta l=0.380\times l_{0}$

Length after expansion is 12 cm

We need to calculate the original length

Using formula of length

$l=l_{0}+\Delta l$

Put the value into the formula

$I=l_{0}+0.380\times l_{0}$

$l=1.38l_{0}$

$l_{0}=\dfrac{l}{1.38}$

$l_{0}=\dfrac{12\times10^{-2}}{1.38}$

$l_{0}=0.0869\ m$

Hence, (a). The magnitude of the force is 0.38416 N.

(b). The original length is 0.0869 m.

4 0

3 years ago

(i). A ball of mass 1.500 kg is attached to the end of a cord 1.50 m long. The ball moves in a horizontal circle. If the cord ca

Aleks04 [339]

(a) Let $v$ be the maximum linear speed with which the ball can move in a circle without breaking the cord. Its centripetal/radial acceleration has magnitude

$a_{\rm rad} = \dfrac{v^2}R$

where $R$ is the radius of the circle.

The tension in the cord is what makes the ball move in its plane. By Newton's second law, the maximum net force on it is

$F = (1.500\,\mathrm{kg}) a_{\rm rad}$

so that

$(1.500\,\mathrm{kg}) \dfrac{v^2}{1.50\,\rm m} = 64.0\,\mathrm N$

Solve for $v$ :

$v^2 = \dfrac{(64.0\,\mathrm N)(1.50\,\mathrm m)}{1.500\,\rm kg} \\\\ \implies \boxed{v = 8.00 \dfrac{\rm m}{\rm s}}$

(b) The net force equation in part (a) leads us to the relation

$F = \dfrac{mv^2}R \implies v = \sqrt{\dfrac{FR}m}$

so that $v$ is directly proportional to the square root of $R$ . As the radius $R$ increases, the maximum linear speed $v$ will also increase, so the cord is less likely to break if we keep up the same speed.

6 0

1 year ago

What is equal to the kinetic energy of a car with a mass of 0.5t (tonne) if it travels evenly at 70 km/h?

Aleonysh [2.5K]

Formula for kinetic energy is 1/2mv^2 so that answer should most probably be 94521.6J

5 0

2 years ago