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borishaifa [10]
3 years ago
5

How do seatbelts reduce injury during a collision ?

Physics
1 answer:
RoseWind [281]3 years ago
3 0
Among drivers and front-seat passengers, seat belts reduce the risk of death by 45%, and cut the risk of serious injury by 50%. Seat belts prevent drivers and passengers from being ejected during a crash. People not wearing a seat belt are 30 times more likely to be ejected from a vehicle during a crash.
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While moving in, a new homeowner is pushing a box across the floor at a constant velocity. The coefficient of kinetic friction b
tresset_1 [31]

Answer:

64.1°

Explanation:

Coefficient of friction = Horizontally resolved pushing force divided by Moving force

Coefficient of friction is 0.437

Resolving pushing force to horizontal = Moving force * Cos ∅

Therefore, 0.437 = Cos ∅

and ∅ = Arch Cos 0.437 = 64.1°

6 0
3 years ago
PLEASE HELP ME!!!!!!!!
Ray Of Light [21]

Answer:

I believe C

Explanation:

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5 0
3 years ago
Read 2 more answers
An isolated capacitor with capacitance C = 1 µF has a charge Q = 45 µC on its plates.a) What is the energy stored in the capacit
Roman55 [17]

Answer:

a) Energy stored in the capacitor, E = 1.0125 *10^{-3} J

b) Q = 45 µC

c) C' = 1.5 μF

d)  E = 6.75 *10^{-4} J

Explanation:

Capacitance, C = 1 µF

Charge on the plates, Q = 45 µC

a) Energy stored in the capacitor is given by the formula:

E = \frac{Q^2}{2C} \\\\E = \frac{(45 * 10^{-6})^2}{2* 1* 10^{-6}}\\\\E = \frac{2025 * 10^{-6}}{2}\\\\E = 1012.5 *10^{-6}\\\\E = 1.0125 *10^{-3} J

b) The charge on the plates of the capacitor will  not change

It will still remains, Q = 45 µC

c)  Electric field is non zero over (1-1/3) = 2/3 of d

From the relation V = Ed,

The voltage has changed by a factor of 2/3

Since the capacitance is given as C = Q/V  

The new capacitance with the conductor in place, C' = (3/2) C

C' = (3/2) * 1μF

C' = 1.5 μF

d) Energy stored in the capacitor with the conductor in place

E = \frac{Q^2}{2C} \\\\E = \frac{(45 * 10^{-6})^2}{2* 1.5* 10^{-6}}\\\\E = \frac{2025 * 10^{-6}}{3}\\\\E = 675 *10^{-6}\\\\E = 6.75 *10^{-4} J

4 0
3 years ago
calculate the resistance of 50m length of wire of cross sectional area 0.01 square mm and of resistivity 5*10 to the power minus
Pachacha [2.7K]
Resistance R = resistivity * length / area

R = 5  10^{-8} * 50 m / 0.01 *  10^{-3} * 10^{-3}

R =  250 Ohms

8 0
3 years ago
A prediction or educated guess that can be tested is ?
Ksivusya [100]
An educated guess is a “hypothesis”
7 0
3 years ago
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