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Zarrin [17]
3 years ago
6

A car has an initial velocity of 11.2 m /sec. the car accelerates at 10.0 m /s2 for 8.0 seconds. what is the velocity of the car

at the end of the 8.0 seconds?
Physics
1 answer:
Vinil7 [7]3 years ago
7 0
Initial velocity(u) = 11.2 m/s.
Final velocity(v) = ?
acceleration(a) = 10.2 m/s²

Using kinematic equation v = u + at

v = 11.2 + 10 x 8 = 11.2 + 80 = 91.2 m/s.

Therefore final velocity is 91.2 m/s.
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How much energy (in kW-h) does a 900 Watt stove use in a week if it is used for 1.5 hours each day?
ioda

Answer:

9.45 kWh

Explanation:

Energy = Power × time

E = 900 W × (1.5 h/day × 7 day)

E = 9450 Wh

E = 9.45 kWh

3 0
3 years ago
A woman exerts a horizontal force of 113 N on a crate with a mass of 31.2 kg.
sergeinik [125]

Answer:

a) 113N

b) 0.37

Explanation:

a) Using the Newton's second law:

\sum Fx =ma

Since the crate doesn't move (static), acceleration will be zero. The equation will become:

\sum Fx = 0

\sumFx = Fm - Ff = 0.

Fm is the applied force

Ff is the frictional force

Since Fm - Ff = 0

Fm = Ff

This means that the applied force is equal to the force of friction if the crate is static.

Since applied force is 113N, hence the magnitude of the static friction force will also be 113N

b) Using the formula

Ff = nR

n is the coefficient of friction

R is the reaction = mg

R = 31.2 × 9.8

R = 305.76N

From the formula

n = Ff/R

n = 113/305.76

n = 0.37

Hence the minimum possible value of the coefficient of static friction between the crate and the floor is 0.37

8 0
2 years ago
A charged particle having mass 6.64 x 10-27 kg (that of a helium atom) moving at 8.70 x 105 m/s perpendicular to a 1.30-T magnet
Fiesta28 [93]

Answer:

the charge of the particle is 2.47 x 10⁻¹⁹ C

Explanation:

Given;

mass of the particle, m = 6.64 x 10⁻²⁷ kg

velocity of the particle, v = 8.7 x 10⁵ m/s

strength of the magnetic field, B = 1.3 T

radius of the circle, r = 18 mm = 1.8 x 10⁻³ m

The magnetic force experienced by the charge is calculated as;

F = ma = qvB

where;

q is the charge of the particle

a is the acceleration of the charge in the circular path

a = \frac{v^2}{r} \\\\ma = qvB\\\\q = \frac{ma}{vB} \\\\q = \frac{mv^2}{rvB} = \frac{mv}{rB} \\\\q = \frac{(6.64\times 10^{-27} ) \times (8.7\times 10^5)}{(1.8\times 10^{-2}) \times (1.3)} \\\\q = 2.47 \ \times 10^{-19} \ C

Therefore, the charge of the particle is 2.47 x 10⁻¹⁹ C

6 0
3 years ago
HELP!
dimaraw [331]
The radiations detected by Arno Penzias and Robert Wilson
were the original heat from the Big Bang.

Today, we call those waves the "Cosmic Microwave Background".

I'm not sure, but I think those guys were awarded a Nobel Prize in Physics
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7 0
3 years ago
Read 2 more answers
a hot piece of copper is placed in an insulated cup. what is the final temperature of the water and copper?
slavikrds [6]

Answer:

Option C. 30°C.

Explanation:

The following data were obtained from the question:

Mass of water (Mw) = 0.5 Kg

Specific heat capacity of water (Cw) = 4.18 KJ/Kg°C

Initial temperature of water (Tw1) =

22°C

Change in temperature (ΔT) = T2 – Tw1 = T2 – 22°C

Mass of copper (Mc) = 0.5 Kg

Specific heat capacity of copper (Cc) = 0.386 KJ/kg°C

Initial temperature of copper (Tc1) = 115°C

Change in temperature (ΔT) = T2 – Tc1 = T2 – 115°C

Final temperature (T2) =..?

Note: Both the water and the piece copper will have the same final temperature and the heat will be zero since the water will cool the piece of copper.

Thus, we can determine the final temperature of the water and copper as follow:

Q = MwCwΔT + McCcΔT

0 = 0.5 x 4.18 x (T2 – 22) + 0.5 x 0.386 x (T2 – 115)

0 = 2.09 (T2 – 22) + 0.193 (T2 – 115)

0 = 2.09T2 – 45.98 + 0.193T2 – 22.195

Collect like terms

2.09T2 + 0.193T2 = 45.98 + 22.195

2.283T2 = 68.175

Divide both side by the coefficient of T2 i.e 2.283

T2 = 68.175/2.283

T2 = 29.8 ≈ 30°C

Therefore, the final temperature of water and copper is 30°C.

8 0
2 years ago
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