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Lapatulllka [165]
3 years ago
15

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

Physics
1 answer:
Talja [164]3 years ago
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

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What is the ACCELERATION of a 50 kg object pushed with a force of 500 newtons *
Anon25 [30]

Answer:

The answer is

<h2>10 m/s²</h2>

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

<h3>10 m/s²</h3>

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
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