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2 years ago

Your friend who has a mass of 80 kg runs with a velocity of 2

Physics

1 answer:

GuDViN [60]2 years ago

5 0

Explanation:

Momentum before = momentum after

m₁ u₁ + m₂ u₂ = m₁ v₁ + m₂ v₂

After she jumps on the skateboard, they have the same velocity, so v₁ = v₂ = v.

m₁ u₁ + m₂ u₂ = (m₁ + m₂) v

The skateboard is originally stationary, so u₂ = 0.

m₁ u₁ = (m₁ + m₂) v

Plugging in values and solving:

(80 kg) (2 m/s) = (80 kg + 2 kg) v

v = 1.95 m/s

Round as needed.

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What is the mass of an object that creates 33,750 joules of energy by traveling at 30 m/sec?

nikklg [1K]

The Energy is Kinetic Energy.

Kinetic Energy = 1/2*mv², Where m is mass in kg, v is velocity in m/s

Energy is 33750 Juoles, v = 30m/s

1/2*mv² = E

1/2*m*30² = 33750

m = (2*33750) / (30²) Using a calculator

m = 75 kg

Mass of object is 75 kg.

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2 years ago

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If a 110 kg go-cart traveling at a velocity of 13.41 m/s has a collision with an impulse of 615 Nxs, what is the

mafiozo [28]

Answer:

5.59 m/s

Explanation:

We are given;

Mass = 110 kg

Initial velocity: u = 13.41 m/s

Force = 615 N

Time(t) = 1 s

Now, the formula for force is;

Force = mass x acceleration

Thus;

615 = 110 × acceleration

\Acceleration(a) = 615/110 = 5.591 m/s²

Now, using Newton's first law of motion, we can find acceleration (a). Thus;

v = u + at

v = 13.41 + (5.591 × 1)

v ≈ 19 m/s

So,the change in velocity is;

Final velocity(v) - Initial velocity(u) = 19 - 13.41 = 5.59 m/s

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2 years ago

A 2 000-kg sailboat experiences an eastward force of 3 000 N by the ocean tide and a wind force against its sails with a magnitu

Vesnalui [34]

Answer:

The magnitude of the resultant acceleration is 2.2 $m/s^2$

Explanation:

Mass (m) of the sailboat = 2000 kg

Force acting on the sailboat due to ocean tide is $F_1$ = 3000N

Eastwards means takes place along the positive x direction

Then $F_{1x}$ = 3000N and $F_{1y}$ = 0

Wind Force acting on the Sailboat is $F_2$ = 6000N directed towards the northwest that means at an angle 45 degree above the negative x axis

Then

$F_{2x}$ = -(6000N) cos 45 degree = -4242.6 N

$F_{2y}$ = (6000N) cos 45 degree = 4242.6 N

Hence , the net force acting on the sailboat in x direction is

$F_x = F_{1x}+ F_{2x}$

= - 3000 N + 4242.6 N

= - 3000 N +4242.6 N

= 1242.6N

Net Force acting on the sailboat in y direction is

$F_y = F_{1y}+ F_{2y}$

= 0+ 4242.6N

= 4242.6N

The magnitude of the resultant force =

Using pythagorean theorm of 1243 N and 4243 N

$\sqrt{(1242.6)^2 + (4242.6)^2$

$\sqrt{(1544054.76) + (17999654.8)}$

$\sqrt{(19543709.5)^2}$

4420.8 N

F = ma

$a = \frac{F}{m}$

$a =\frac{4420.8}{ 2000}$

=2.2 $m/s^2$

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2 years ago

Describe electrons.<br> Location:<br> Charge:<br> Mass

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A proton starts from rest and gains 8.35 x 10^-14 joule of kinetic energy as it accelerates between points A and B in an electri

antoniya [11.8K]

Answer:

5.22 x 10^5 V

Explanation:

guessed on castle learning and got it right

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