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1 year ago

What is the horsepower of a 1,500 kg car that can go to the top of a 360 m high hill in exactly 1 min?

Physics

1 answer:

Airida [17]1 year ago

7 0

Answer:

W = m g h work that must be done on car

P = W / t power that must be input (in Watts)

P = m g h / t = 1500 kg * 9.8 m/s^2 * 360 m / 60 sec

P = 88,200 watts

P = 88,200 watts / 746 watts / hp = 118 hp

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Pls help me guys simple question

Airida [17]

Answer:

please write neater

Explanation:

can you write neater so I can answer th question but also is a equal to b

7 0

3 years ago

The cardiovascular system consists of

guapka [62]

Consists of A. good luck

3 0

2 years ago

if a star 100 light years from earth is beginning to expand into a giant star how long will it take for astronomers to observe t

VikaD [51]

Answer:

100years later

Explanation:

Because the lights will arrive at world after 100 years later.

7 0

2 years ago

The force of attraction between a -165.0 uC and +115.0 C charge is 6.00 N. What is the separation between these two charges in m

Simora [160]

Answer:

The distance between the charges is 5,335.026 m

Explanation:

To obtain the forces between the particles, we can use Coulomb's Law in scalar form, this is, the force between the particles will be:

$F = k \frac{q_1 q_2}{d^2}$

where k is Coulomb's constant, $q_1$ and $q_2$ are the charges and d is the distance between the charges.

Working a little the equation, we can take:

$d^2 = k \frac{q_1 q_2}{F}$

$d = \sqrt{ k \frac{q_1 q_2}{F}}$

And this equation will give us the distance between the charges. Taking the values of the problem

$k= 9.00 \ 10^9 \frac{N \ m^2}{C^2} \\q_1 = 165.0 \mu C \\q_2 = 115.0 C\\F=- 6.00$

(the force has a minus sign, as its attractive)

$d = \sqrt{ 9.00 \ 10^9 \frac{N \ m^2}{C^2} \frac{(165.0 \mu C) (115.0 C)}{- 6.00 \ N}}$

$d = \sqrt{ 28,462,500 \ m^2}}$

$d = 5,335.026 m$

And this is the distance between the charges.

3 0

3 years ago

To warm 2.0 l of tea (d = 1.01 g/ml; sp. heat = a cook places a 500 g block of stone at a temperature of 200f into the teapot. a

IRISSAK [1]

Volume of tea V = 2.0L = 2000 mL density d = 1.01 g/ mL mass of tea m = V * d = 2000mL * 1.01g/mL = 2020 gWhen we assume that the tea was initially at 72, the final temperature of the tea in F is 91.
The answer in this question is B. 91

4 0

3 years ago