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3 years ago

Suppose you give a 10 Newton push to Ryan on skis (he weighs 50 kg), how much will he accelerate?

1 answer:

3 0

Well we can just use F=ma. The force is 10N, the mass is 50 kg, solve for a. Well since we kg and N, no conversion is necessary. So just plugging in the numbers, we get

10N = 50 kg · a

$\frac{10N}{50kg}=a$

A newton is just $\frac{kg·m}{s^{2}}$

$a=\frac{\frac{10kg·m}{s^{2}}}{50kg}$

The s^2 and 50 kg you multiply

$a=\frac{10kg·m}{50kg·s^{2}}$

The kg's cancel and 10/50 is 1/5

$\frac{1}{5}·\frac{m}{s^{2}}$

So the acceleration is 1/5 m/s^2

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A tall cylinder contains 30 cm of water. Oil is carefully poured into the cylinder, where it floats on top of the water, until t

Anastaziya [24]

Answer:

The gauge pressure is $P_g = 2058 \ P_a$

Explanation:

From the question we are told that

The height of the water contained is $h_w = 30 \ cm = 0.3 \ m$

The height of liquid in the cylinder is $h_t = 40 \ cm = 0.4 \ m$

At the bottom of the cylinder the gauge pressure is mathematically represented as

$P_g = P_w + P_o$

Where $P_w$ is the pressure of water which is mathematically represented as

$P_w = \rho_w * g * h_w$

Now $\rho_w$ is the density of water with a constant values of $\rho_w = 1000 \ kg /m^3$

substituting values

$P_w = 1000 * 9.8 * 0.3$

$P_w = 2940 \ Pa$

While $P_o$ is the pressure of oil which is mathematically represented as

$P_o = \rho_o * g * (h_t -h_w )$

Where $\rho _o$ is the density of oil with a constant value

$\rho _o = 900 \ kg / m^3$

substituting values

$P_o = 900 * 9.8 * (0.4 - 0.3)$

$P_o = 882 \ Pa$

Therefore

$P_g = 2940 - 882$

$P_g = 2058 \ P_a$

6 0

3 years ago

La energía cinética es la energía que presentan los cuerpos que se encuentran en movimiento.

myrzilka [38]

Do you speak a little English cuz I can’t help you if a can’t understand you

6 0

3 years ago

Where does most of the energy we use at home/our cars come from?

KengaRu [80]

It comes from the sun, and then it is converted to energy/electricity (by solar panels)

4 0

3 years ago

What is the ACCELERATION of a 50 kg object pushed with a force of 500 newtons *

Anon25 [30]

Answer:

The answer is

Explanation:

To find the acceleration of an object given the force and mass we use the formula

<h3> $acceleration = \frac{force}{mass}$ </h3>

From the question

mass of object = 50 kg

force = 500 N

So the acceleration is

<h3> $acceleration = \frac{500}{50} \\ = \frac{50}{5}$ </h3>

We have the final answer as

Hope this helps you

5 0

3 years ago

A spring with force constant of 59 N/m is compressed by 1.3 cm in a hockey game machine. The compressed spring is used to accele

Furkat [3]

Answer:

The puck moves a vertical height of 2.6 cm before stopping

Explanation:

As the puck is accelerated by the spring, the kinetic energy of the puck equals the elastic potential energy of the spring.

So, 1/2mv² = 1/2kx² where m = mass of puck = 39.2 g = 0.0392 g, v = velocity of puck, k = spring constant = 59 N/m and x = compression of spring = 1.3 cm = 0.013 cm.

Now, since the puck has an initial velocity, v before it slides up the inclined surface, its loss in kinetic energy equals its gain in potential energy before it stops. So

1/2mv² = mgh where h = vertical height puck moves and g = acceleration due to gravity = 9.8 m/s².

Substituting the kinetic energy of the puck for the potential energy of the spring, we have

1/2kx² = mgh

h = kx²/2mg

= 59 N/m × (0.013 m)²/(0.0392 kg × 9.8 m/s²)

= 0.009971 Nm/0.38416 N

= 0.0259 m

= 2.59 cm

≅ 2.6 cm

So the puck moves a vertical height of 2.6 cm before stopping

3 0

3 years ago