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2 years ago

Please help me! Will give brainliest.

Physics

2 answers:

AleksAgata [21]2 years ago

6 0

Answer:

B. Boats need to float.

maxonik [38]2 years ago

5 0

It’s B. because the density between the water and the boat will cause it to float even if it has a little bit weight on it.

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A mass of 5 kg stretches a spring 20 cm. The mass is acted on by an external force of 10 sin t 6 N (newtons) and moves in a medi

nordsb [41]

Answer:

u" + 40u' + 49u = 2 sin(t/6)

upp + 40up + 49u = 2 sin(t/6)

Explanation:

Step 1: Data given

mass = 5 kg

L = 20 cm = 0.2 m

F = 10 sin(t/6)N

Fd(t) = - 6 N

u(0) = 0.03 m/s

u(0) = 0

u'(0) = 3 cm/s

Step 2:

ω =kL

k = ω/L = m*g /L = (5*9.8)/0.2 = 245 kg/s²

Since Fd(t) = -γu'(t) we know:

γ =- Fd(t) / u'(t) = 6N/ 0.03 m/s = 200 Ns/m

The initial value problem which describes the motion of the mass is given by

5u" + 200u' + 245u = 10 sin(t/6) u(0) = 0 ; u'(0) = 0.03

This is equivalent to:

u" + 40u' + 49u = 2 sin(t/6) u(0) = 0 ; u'(0) = 0.03

upp + 40up + 49u = 2 sin(t/6)

With u in m and t in s

5 0

3 years ago

A uniform disk with a mass of 5.0 kg and diameter 30 cm rotates on a frictionless fixed axis through its center and perpendicula

igomit [66]

Answer:

Angular acceleration of the disk will be $\alpha =10.714rad/sec^2$

Explanation:

We have given mass of the disk m = 5 kg

Diameter of the disk d = 30 cm = 0.3 m

So radius $r=\frac{d}{2}=\frac{0.3}{2}=0.15m$

Moment of inertia of disk is given by $I=\frac{1}{2}mr^2=\frac{1}{2}\times 5\times 0.15^2=0.056kgm^2$

Force is given by F=4 N

Torque is given as $\tau =Fr=4\times 0.15=0.6N-m$

We also know that torque is given by $\tau =I\alpha$

$0.6=0.056\times \alpha$

$\alpha =10.714rad/sec^2$

5 0

3 years ago

Two identical bowling balls are rolling on a horizontal floor without slipping. The initial speed of both balls is v = 10 m/s. B

Lapatulllka [165]

Answer:

The difference between frictionless ramp and a regular ramp is that on a frictionless ramp the ball cannot roll it can only slide, but on a regular ramp the ball can roll without slipping.

We will use conversation of energy.

$K_A_1 + U_A_1 = K_A_2 + U_A_2\\\frac{1}{2}I\omega^2 + \frac{1}{2}mv^2 + 0 = 0 + mgH_A$

Note that initial potential energy is zero because the ball is on the bottom, and the final kinetic energy is zero because the ball reaches its maximum vertical distance and stops.

For the ball B;

$K_B_1 + U_B_1 = K_B_2 + U_B_2$

$\frac{1}{2}I_B\omega^2 + \frac{1}{2}mv^2 + 0 = 0 + mgH_B$

The initial velocities of the balls are equal. Their maximum climbing point will be proportional to their final potential energy. Since their initial kinetic energies are equal, their final potential energies must be equal as well.

Hence, both balls climb the same point.

Explanation:

4 0

3 years ago

T-Joe (65 kg) is running at 3 m/s. T-Brud (50 kg) is running at 4 m/s. What would be T-Joe's momentum?

Debora [2.8K]

Answer:

$P_J=195N$