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Aliun [14]
3 years ago
11

what is the average acceleration of a car if a car rest from 0m/s and reaches a final velocity is 50m/s in 10m/s

Physics
1 answer:
GuDViN [60]3 years ago
6 0

Answer:

Assuming that you meant the final velocity of 50 m/s was reached in 10 s, the answer would be 5 m/s^2.

Explanation:

V_{f} = V_{i} + at

So we update that with the values that we have.

50 = 0 +a(10)

then simplify that using algebra to solve for a and we get 5 m/s^2

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A runner generates 1260 W of thermal energy. If this heat has to be removed only by evaporation, how much water does this runner
sergey [27]

Answer:0.502kg

Explanation:

F4om the relation

Power x time = mass x latent heat of vapourization

P.t=ML

1260 * 15 *60 = M * 22.6 * 10^5

M= 1134000/(22.6 *10^5)

M=0.502kg=502g

3 0
3 years ago
If a point has 40 J of energy and the electric potential is 8 V, what must be the charge?
Alekssandra [29.7K]

If a point has 40 J of energy and the electric potential is 8 V, the charge must be: A. 5 C

<u>Given the following the details;</u>

  • Energy = 40 Joules
  • Electric potential = 8 Volts

To find the quantity of charge;

Mathematically, the quantity of charge with respect to electric potential is given by the formula;

Quantity \; of \; charge = \frac{Energy}{Electric \; potential}

Substituting the values into the formula, we have;

Quantity \; of \; charge = \frac{40}{8}

<em>Quantity of charge = 5 Coulombs</em>

Therefore, the quantity of charge must be <em>5 Coulombs.</em>

Find more information: brainly.com/question/21808222

8 0
3 years ago
Read 2 more answers
Two resistances, R1 and R2, are connected in series across a 9-V battery. The current increases by 0.450 A when R2 is removed, l
Rina8888 [55]

Answer:

a. R1 = 0.162 Ω

b. R2 = 0.340 Ω

Explanation:

Since the resistors R1 and R2 are connected in series, the current flowing through them when the 9 V battery is applied is 9/R1 + R2.

When the current increases by 0.450 A wen only R1 is in the circuit, the current is

9/R1 + R2 + 0.450 A = 9/R1       (1)

When the current increases by 0.225 A when only R2 is in the circuit, the current is

9/R1 + R2 + 0.225 A = 9/R2       (2)

equation (1) - (2) equals

9(1/R1 - 1/R2) = 0.450 A - 0.225

9(1/R1 - 1/R2) = 0.125

(1/R1 - 1/R2) = 0.125 A/9 = 0.0138

1/R1 = 0.0138 + 1/R2

R1 = R2/(1 + 0.0138R2)     (3)

From (1)

9/R1 - 9/R1 + R2 = 0.450 A

9R2/[R1(R1 + R2)] = 0.450 A

R2/[R1(R1 + R2)] = 0.450 A/9 = 0.5

R2/[R1(R1 + R2)] = 0.5    (4)

From (3) R2/R1 = (1 + 0.0138R2) and from (4) R2/R1 = 0.5(R1 + R2). So,

(1 + 0.0138R2) = 0.5(R1 + R2)

0.5R1 + 0.5R2 = 1 + 0.0138R2

0.5R1 = 1 + 0.0138R2 - 0.5R2

0.5R1 = 1 - 0.4862R2        (5)

Substituting (3) into (5) we have

0.5R2/(1 + 0.0138R2) = 1 - 0.4862R2

R2 = (1 + 0.0138R2)(1 - 0.4862R2)

R2 = 1 - 0.4724R2 - 0.0067R2²

Collecting like terms, we have

0.0067R2² + 0.4724R2 + R2 - 1 = 0

0.0067R2² + 1.4724R2 - 1 = 0

Using the quadratic formula,

R_{2} = \frac{-1.4724 +/-\sqrt{(1.4724)^{2} - 4 X 0.0067 X -1} }{2 X 0.0067}  \\= \frac{-1.4724 +/-\sqrt{2.1680 + 0.0268} }{0.0268}\\= \frac{-1.4724 +/-\sqrt{2.1948} }{0.0268}\\= \frac{-1.4724 +/- 1.4815 }{0.0268}\\= \frac{-1.4724 + 1.4815 }{0.0268} or \frac{-1.4724 - 1.4815 }{0.0268}\\= \frac{0.0091 }{0.0268} or \frac{-2.9539}{0.0268}\\= 0.340 or -110.22

We choose the positive answer.

So R2 = 0.340 Ω

From (5)

R1 = 0.5 - 0.9931R2

   = 0.5 - 0.9931 × 0.340

   = 0.5 - 0.338

   = 0.162 Ω

a. R1 = 0.162 Ω

b. R2 = 0.340 Ω

5 0
3 years ago
The image of an object formed by plane mirror is​
m_a_m_a [10]
Answer of your question is in this photo

8 0
2 years ago
At which point does the planet have the least gravitational force acting on it?
Elza [17]

Answer:

At which point does the planet have the least gravitational force acting on it?

Explanation:

In an elliptical orbit, when a planet is at its furthest point from the Sun, it is under the least amount of gravity, meaning that the force of gravity is strongest when it is closest.

5 0
3 years ago
Read 2 more answers
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