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Len [333]
2 years ago
8

In physics, what is the positive and negative sign used for when applied to velocity and acceleration?

Physics
1 answer:
Kobotan [32]2 years ago
6 0

Answer:

Opposite direction

Explanation:

For example: The first object travels in the direction pointing to east, while the other one travels pointing to the west.

The first object will have a positive value while the second one will have a negative, cause it's going to the opposite direction

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Difference between rest and motion​
creativ13 [48]
Rest - it is the state in which body doesn’t move from it’s place

motion - it is the state in which body moves from it’s place
3 0
3 years ago
Properties of light that define light as a wave?
MAVERICK [17]
The first choices are correct, because the second choices could happen by things other than light.
6 0
2 years ago
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Help me please, please !
steposvetlana [31]

Answer:

b and d

a, c, e, and f

Explanation:

Ideal gas law:

PV = nRT

Solving for temperature:

T = PV / (nR)

Therefore, temperature is directly proportional to pressure and volume, and inversely proportional to the number of molecules.

T = k PV / N

Let's say that T₀ is the temperature when P = 100 kPa, V = 4 L, and N = 6×10²³.

a) T = k PV / N = T₀

b) T = k (2P) V / N = 2T₀

c) T = k (P/2) (2V) / N = T₀

d) T = k PV / (N/2) = 2T₀

e) T = k P (V/2) / (N/2) = T₀

f) T = k (P/2) V / (N/2) = T₀

b and d have the highest temperature,

a, c, e, and f have the lowest temperature.

8 0
3 years ago
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An electric bulb is marked 40volts ,230w another bulb is marked 40w,110v
Andrej [43]

Answer:

a. The ratio of their resistance is 2783:64

b. The ratio of their energy is 4:23

c. The charge on the first bulb is 5.75 C

The charge on the second bulb is 0.\overline {36} C

Explanation:

The voltage on one of the electric bulbs, V₁ = 40  volts

The power rating of the bulb, P₁ = 230 w

The voltage on the other electric bulbs, V₂ = 110 volts

The power rating of the bulb, P₂ = 40 w

a. The power is given by the formula, P = I·V = V²/R

Therefore, R = V²/P

For the first bulb, the resistance, R₁ = 40²/230 ≈ 6.96

The resistance of the second bulb, R₂ = 110²/40

The ratio of their resistance, R₂/R₁ = (110²/40)/(40²/230) = 2783/64

∴ The ratio of their resistance, R₂:R₁ = 2783:64

b. The energy of a bulb, E = t × P

Where;

t = The time in which the bulb is powered on

∴ The energy of the first bulb, E₁ = 230 w × t

The energy of the second bulb, E₂ = 40 w × t

The ratio of their energy, E₂/E₁ = (40 w × t)/(230 w × t) = 4/23

∴ The ratio of their energy, E₂:E₁ = 4:23

c. The charge on a bulb, 'Q', is given by the formula, Q = I × t

Where;

I = The current flowing through the bulb

From P = I·V, we get;

I = P/V

For the first bulb, the current, I = 230 w/40 V = 5.75 amperes

The charge on the first bulb per second (t = 1) is therefore;

Q₁ = 5.75 A × 1 s = 5.75 C

The charge on the first bulb, Q₁ = 5.75 C

Similarly, the charge on the second bulb, Q₂ = (40 W/110 V) × 1 s = 0.\overline {36} C

The charge on the second bulb, Q₂ = 0.\overline {36} C.

d. The question has left out parts

4 0
2 years ago
A sled slides along a horizontal surface on which the coefficient of kinetic friction is 0.25. Its velocity at point A is 8.6 m/
jenyasd209 [6]

Answer:

\Delta t =1.31\ s

Explanation:

given,

coefficient of kinetic friction, μ = 0.25

Speed of sled at point A = 8.6 m/s

Speed of sled at point B = 5.4 m/s

time taken to travel from point A to B.

we know,

J = F Δ t

J is the impulse

where  F is the frictional force.

t is the time.

we also know that impulse is equal to change in momentum.

J = m(v_f - v_i)

frictional force

F = μ N

where as N is the normal force

now,

F\Delta t = m(v_f -v_i)

\mu m g \times \Delta t = m(v_f-v_i)

\mu g \times \Delta t = v_f-v_i

\Delta t =\dfrac{v_f-v_i}{\mu g}

\Delta t =\dfrac{8.6-5.4}{0.25\times 9.8}

\Delta t =1.31\ s

time taken to move from A to B is equal to 1.31 s

3 0
3 years ago
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