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Delicious77 [7]
3 years ago
11

Assuming 84.0% efficiency for the conversion of electrical power by the motor, what current must the 13.0-V batteries of a 716 k

g electric car be able to supply to climb a 3.00 x 102 m high hill in 2.00 min at a constant 22.0 m/s speed while exerting 7.00 x 102 N of force to overcome air resistance and friction
Physics
1 answer:
tino4ka555 [31]3 years ago
8 0

Answer:

\mathbf{ current(I) =1766.67 \ A}

Explanation:

Given that:

The air resistance and friction = 700 N

The gravity caused force = 716 × 9.8 = 7016.8

Total force = (7016.8 + 700) N

Total force = 7716.8 N

∴

13 \times  current(I) \times 0.84 = \dfrac{7716.8 \times 300}{2 \times 60}

current(I) \times 10.92= 19292

current(I) = \dfrac{19292}{10.92}

\mathbf{ current(I) =1766.67 \ A}

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zimovet [89]
Static frictional force = ƒs = (Cs) • (Fɴ)
              2.26 = (Cs) • m • g
              2.26 = (Cs) • (1.85) • (9.8)
           Cs = 0.125 
kinetic frictional force = ƒκ = (Cκ) • (Fɴ)
              1.49 = (Cκ) • m • g
              1.49 = (Cκ) • (1.85) • (9.8)
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3 0
3 years ago
A car starts from rest and accelerates uniformly at a rate of 2.0 meter per second squared for 4.0 seconds. During this time int
vagabundo [1.1K]

Answer:

16

Explanation:

\frac{1}{2}  \times 2 \times  {4}^{2}  = 16

7 0
3 years ago
A coin of mass m rests on a turntable a distance r from the axis of rotation. The turntable rotates with a frequency of f. What
Crank

Answer:

\mu = \frac{r (2\pi f)^{2}}{g}

Explanation:

N = normal force acting on the coin

Normal force in the upward direction balances the weight of the coin, hence

N = mg

f = frequency of rotation

Angular velocity of turntable is hence given as

w = 2\pi f

r = distance from the axis of rotation

\mu = minimum coefficient of static friction

static frictional force is given as

f = \mu N\\f = \mu mg

The  static frictional force provides the necessary centripetal force , hence

Centripetal force = Static frictional force

m r w^{2} = \mu mg\\r w^{2} = \mu g\\\\\mu = \frac{r w^{2}}{g} \\\mu = \frac{r (2\pi f)^{2}}{g}

3 0
3 years ago
Here, mA = 3.65 kg and mB = 7.05 kg. The string connecting the two objects is of negligible mass and the pulley is frictionless.
mr_godi [17]

Answer:

The magintude of the acceleration for both objects is 3.11m/s^2

Explanation:

Drawing a free body diagram on the two boxes we can analyze the system more easily.

we can take the acceleration going up as positive for reference purposes.

for mA let's suppose that is ascending so:

T_A-m_A*g=m_A*a

and for mB (descending):

T_B-m_B*g=-m_B*a

T_A=T_B

because the two boxes has the same acceleration because they are attached together:

m_B*g-m_B*a-m_a*g=m_a*a\\(m_B-m_A)*g=(m_a+m_B)*a\\a=\frac{(7.05kg-3.65kg)*9.8m/s^2}{3.65kg+7.05kg}\\\\a=3.11m/s^2

So the magintude of the acceleration for both objects is 3.11m/s^2

5 0
3 years ago
If a cup of coffee is at 90°C and a person with a body temperature of 36'C touches it,how will heat flow between them
Arlecino [84]

Answer:

the heat always transfers from high temperature to low temperature body without aid of any external energy to this law the heat transfers from cup of coffee to the person body until both bodies reaches to the equilibrium state    

Explanation:

5 0
3 years ago
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