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BlackZzzverrR [31]
3 years ago
12

What is the static friction force??

Physics
1 answer:
Paul [167]3 years ago
6 0
55 Kg has a weight of 55x9.8= 539 N
That is equal to the Normal force.
The static friction = 0.19 x 539 = 102.4 N
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Which of the graphs describes the motion of a person who first rode her bicycle at constant speed and then rested?
sergeinik [125]

Answer:

Its graph 1

Explanation:

She started at the origin and kept riding her bike until she stopped which causes the line to go staright because she's not moving.

5 0
2 years ago
The driver of a 1750 kg car traveling on a horizontal road at 110 km/h suddenly applies the brakes. Due to a slippery pavement,
Alexeev081 [22]

Answer: a=-2.4525 m/s^2

d=s=190.3 m

Explanation:The only force that is stopping the car and causing deceleration is the frictional force Fr

Fr = 25% of weight

W=mg

W=1750*9.81

W=17167.5

Hence

Fr=\frac{25}{100} * -17167.5\\\\Fr=-4291.875 N

Frictional force is negative as it acts in opposite direction

According to newton second law of motion

F=ma

hence

a=Fr/m

a=-4291.875/1750\\a=-2.4525

given

u= 110 km/h

u=110*1000/3600

u=30.55 m/s

to get t we know that final velocity v=0

v^2=u^2+2as\\0=30.55^2-2*2.4525*s\\s=190.34m

3 0
3 years ago
A spaceship whose rest length is 350m has a speed of .82c
igomit [66]

Answer:

t'=1.1897*10^{-6} s

t'=1.1897 μs

Explanation:

First we will calculate the velocity of micrometeorite relative to spaceship.

Formula:

u=\frac{u'+v}{1+\frac{u'*v}{c^{2}}}

where:

v is the velocity of spaceship relative to certain frame of reference =  -0.82c (Negative sign is due to antiparallel track).

u is the velocity of micrometeorite relative to same frame of reference as spaceship = .82c (Negative sign is due to antiparallel track)

u' is the relative velocity of micrometeorite with respect to spaceship.

In order to find u' , we can rewrite the above expression as:

u'=\frac{v-u}{\frac{u*v}{c^{2} }-1 }

u'=\frac{-0.82c-0.82c}{\frac{0.82c*(-0.82c)}{c^{2} }-1 }

u'=0.9806c

Time for micrometeorite to pass spaceship can be calculated as:

t'=\frac{length}{Relatie seed (u')}

t'=\frac{350}{0.9806c}     (c = 3*10^8 m/s)

t'=\frac{350}{0.9806* 3.0*10^{8} }

t'=1.1897*10^{-6} s

t'=1.1897 μs

4 0
3 years ago
Ignoring the electron spin what is the largest possible energy difference, if the magnetic field is 2.02 tesla?
Nonamiya [84]

Complete Question:

When specially prepared Hydrogen atoms with their electrons in the 6f state are placed into a strong uniform magnetic field, the degenerate energy levels split into several levels. This is the so called normal Zeeman effect.

Ignoring the electron spin what is the largest possible energy difference, if the magnetic field is 2.02 Tesla?

Answer:

ΔE = 1.224 * 10⁻²² J

Explanation:

In the 6f state, the orbital quantum number, L = 3

The magnetic quantum number, m_{L} = -3, -2, -1, 0, 1, 2, 3

The change in energy due to Zeeman effect is given by:

\triangle E = m_{L} \mu_{B} B

Magnetic field B = 2.02 T

Bohr magnetron, \mu_{B} = 9.274 * 10^{-24} J/T

\triangle E = 6 * 9.274 * 10^{-24} * 2.2\\

ΔE = 1.224 * 10⁻²² J

5 0
3 years ago
If you are traveling at 75 km/h how long will it take to travel 32 km?
Oksi-84 [34.3K]

Answer:

This would be 24 minutes

Explanation:

7 0
3 years ago
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