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3 years ago

11

A motorcycle drives up a steeply inclined ramp. The work done on the motorcycle by the Earthâ€™s gravitational force is ____. 1.

positive 2. negative 3. zero

1 answer:

Leto [7]3 years ago

8 0

Answer:

be

Explanation:

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A student produces a power of p = 0.87 kw while pushing a block of mass m = 75 kg on an inclined surface making an angle of Î¸ =

Paul [167]

When block is pushed upwards along the inclined plane

the net force applied on the block will be given as

$F_{net} = mg sin\theta + \mu_k mg cos\theta$

here we know that

m = 75 kg

$\theta = 8.5 degree$

$\mu_k = 0.16$

now plug in all values into this

$F_{net} = 75\times 9.8 sin8.5 + 0.16 \times 75\times 9.8 cos8.5$

$F = 225 N$

now for finding the power is given as

$P = Fv$

$0.87 \times 10^3 = 225 \time v$

$v = \frac{870}{225} = 3.87 m/s$

6 0

3 years ago

Suppose that a spiral galaxy is located at the center of a spherically symmetric dark matter halo

Novosadov [1.4K]

To solve the problem it is necessary to use the concepts of Orbital Speed considering its density, and its angular displacement.

In general terms the Orbital speed is described as,

$V_{orbit} = \sqrt{\frac{G\rho 4\pi r^3}{3}}$

PART A) If the orbital speed of a star in this galaxy is constant at any radius, then,

$\frac{4\pi G\rho r}{3} = \frac{v^2}{r}$

$\frac{4\pi G\rho r}{1} = \frac{3v^2}{r}$

$\frac{\rho}{1} = \frac{3v^2}{r^2 4\pi G}$

$\rho = \frac{1}{r^2}$

PART B) This time we have $v=\omega t$ , where $\omega$ is the angular velocity (constant at this case)

$\frac{4\pi G\rho r}{3} = \frac{v^2}{r}$

$\frac{4\pi G\rho r}{3} = \frac{(\omega r)^2}{r}$

$\rho = \frac{3\omega r}{4\pi Gr}$

$\rho = \frac{3\omega^2}{4\pi G} \propto constant$

PART C) If the total mass interior to any radius r is a constant,

$\frac{4\pi G\rho r}{3} = \frac{v^2}{r}$

$\frac{GM}{r^2}=\frac{v^2}{r}$

$v = \sqrt{\frac{GM}{r}}$

$v= \sqrt{\frac{1}{r}}$

3 0

4 years ago

slader How much energy is required to move a 1040 kg object from the Earth's surface to an altitude four times the Earth's radiu

andrew-mc [135]

Answer:

ΔU = 5.21 × 10^(10) J

Explanation:

We are given;

Mass of object; m = 1040 kg

To solve this, we will use the formula for potential energy which is;

U = -GMm/r

But we are told we want to move the object from the Earth's surface to an altitude four times the Earth's radius.

Thus;

ΔU = -GMm((1/r_f) - (1/r_i))

Where;

M is mass of earth = 5.98 × 10^(24) kg

r_f is final radius

r_i is initial radius

G is gravitational constant = 6.67 × 10^(-11) N.m²/kg²

Since, it's moving to altitude four times the Earth's radius, it means that;

r_i = R_e

r_f = R_e + 4R_e = 5R_e

Where R_e is radius of earth = 6371 × 10³ m

Thus;

ΔU = -6.67 × 10^(-11) × 5.98 × 10^(24)

× 1040((1/(5 × 6371 × 10³)) - (1/(6371 × 10³))

ΔU = 5.21 × 10^(10) J

7 0

3 years ago

A 68.0-kg person jumps from rest off a 2.20-m-high tower straight down into the water. Neglect air resistance during the descent

Margarita [4]

Answer:

$F= 1333.767\,N$

Explanation:

The velocity of the swimmer just before touching the water is:

$v = -\sqrt{2\cdot (9.807\,\frac{m}{s^{2}} )\cdot (2.20\,m)}$

$v \approx 6.569\,\frac{m}{s}$

The average force exerted on the diver by the water is determined by the use of the Principle of Energy Conservation and the Work-Energy Theorem:

$(68\,kg)\cdot (9.807\,\frac{m}{s^{2}})\cdot (2.20\,m) +\frac{1}{2}\cdot (68\,kg)\cdot (6.569\,\frac{m}{s} )^{2}-F\cdot(2.20\,m) = 0\,J$

$F= 1333.767\,N$

6 0

3 years ago

Read 2 more answers

This mathematical model describes the changes that occur in a sample of

Veseljchak [2.6K]

Answer:

the correct answer is B

Explanation:

The kinetic model of the movement describes that the movement of the molecules increases with the increase of their internal energy, which in a macroscopic sample is reflected in an increase in the temperature of the sample.

The sample graph shows the function of temperature over time, which is why our kinetic model establishes that there is an increase in the movement of water molecules.

Consequently the correct answer is B

8 0

3 years ago