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3 years ago

As a 2-kg ball rolls down a ramp, the net force on it is 10 N. What is the acceleration?

1 answer:

4 0

That all depends on the direction of the force. If the force is parallel to the ramp, then the acceleration of the ball is 5 m/s^2. If the force isn't parallel to the ramp, then the acceleration is less.

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Kyle is flying a helicopter at 125 m/s on a heading of 325 o . If a wind is blowing at 25 m/s toward a direction of 240.0 o , wh

frosja888 [35]

Answer:

The resultant velocity of the helicopter is $\vec v_{H} = \left(89.894\,\frac{m}{s}, -93.348\,\frac{m}{s}\right)$ .

Explanation:

Physically speaking, the resulting velocity of the helicopter ( $\vec v_{H}$ ), measured in meters per second, is equal to the absolute velocity of the wind ( $\vec v_{W}$ ), measured in meters per second, plus the velocity of the helicopter relative to wind ( $\vec v_{H/W}$ ), also call velocity at still air, measured in meters per second. That is:

$\vec v_{H} = \vec v_{W}+\vec v_{H/W}$ (1)

In addition, vectors in rectangular form are defined by the following expression:

$\vec v = \|\vec v\| \cdot (\cos \alpha, \sin \alpha)$ (2)

Where:

$\|\vec v\|$ - Magnitude, measured in meters per second.

$\alpha$ - Direction angle, measured in sexagesimal degrees.

Then, (1) is expanded by applying (2):

$\vec v_{H} = \|\vec v_{W}\| \cdot (\cos \alpha_{W},\sin \alpha_{W}) +\|\vec v_{H/W}\| \cdot (\cos \alpha_{H/W},\sin \alpha_{H/W})$ (3)

$\vec v_{H} = \left(\|\vec v_{W}\|\cdot \cos \alpha_{W}+\|\vec v_{H/W}\|\cdot \cos \alpha_{H/W}, \|\vec v_{W}\|\cdot \sin \alpha_{W}+\|\vec v_{H/W}\|\cdot \sin \alpha_{H/W} \right)$

If we know that $\|\vec v_{W}\| = 25\,\frac{m}{s}$ , $\|\vec v_{H/W}\| = 125\,\frac{m}{s}$ , $\alpha_{W} = 240^{\circ}$ and $\alpha_{H/W} = 325^{\circ}$ , then the resulting velocity of the helicopter is:

$\vec v_{H} = \left(\left(25\,\frac{m}{s} \right)\cdot \cos 240^{\circ}+\left(125\,\frac{m}{s} \right)\cdot \cos 325^{\circ}, \left(25\,\frac{m}{s} \right)\cdot \sin 240^{\circ}+\left(125\,\frac{m}{s} \right)\cdot \sin 325^{\circ}\right)$ $\vec v_{H} = \left(89.894\,\frac{m}{s}, -93.348\,\frac{m}{s}\right)$

The resultant velocity of the helicopter is $\vec v_{H} = \left(89.894\,\frac{m}{s}, -93.348\,\frac{m}{s}\right)$ .

8 0

3 years ago

A satellite, orbiting the earth at the equator at an altitude of 400 km, has an antenna that can be modeled as a 1.76-m-long rod

ivann1987 [24]

Answer:

The inducerd emf is 1.08 V

Solution:

As per the question:

Altitude of the satellite, H = 400 km

Length of the antenna, l = 1.76 m

Magnetic field, B = $8.0\times 10^{- 5}\ T$

Now,

When a conducting rod moves in a uniform magnetic field linearly with velocity, v, then the potential difference due to its motion is given by:

$e = - l(vec{v}\times \vec{B})$

Here, velocity v is perpendicular to the rod

Thus

e = lvB (1)

For the orbital velocity of the satellite at an altitude, H:

$v = \sqrt{\frac{Gm_{E}}{R_{E}} + H}$

where

G = Gravitational constant

$m_{e} = 5.972\times 10^{24}\ kg$ = mass of earth

$R_{E} = 6371\ km$ = radius of earth

$v = \sqrt{\frac{6.67\times 10^{- 11}\times 5.972\times 10^{24}}{6371\times 1000 + 400\times 1000} = 7670.018\ m/s$

Using this value value in eqn (1):

$e = 1.76\times 7670.018\times 8.0\times 10^{- 5} = 1.08\ V$

5 0

3 years ago

A 350-N child is in a swing that is attached to a pair of ropes 2.10 m long. Find the gravitational potential energy of the chil

o-na [289]

Answer:

a) U = 735 J , b) U = 125.7 J , c) U = 0 J

Explanation:

The gravitational power energy is

U = mg y - mg y₀

The last value is a constant, for simplicity we can make it zero, if the lowest point is at the origin of the coordinate system, which in this case we will place in the lowest part

a) Rope is horizontal

The height in this case is the same length of the rope

y = 2.10 m

w = mg = 350 N

U = 350 2.10

U = 735 J

b) when the angle is 34º

y = L - L cos 34

y = L (1- cos34)

y = 2.10 (1- cos 34)

y = 0.359 m

U = 350 0.359

U = 125.7 J

c) in this case this point coincides with the reference system

y = 0

U = 0 J

4 0

3 years ago

Which players are usually the tallest on their team, and stay close to the basket so they can shoot and rebound the ball?

alexdok [17]

Answer:

Center

Explanation:

The center is the tallest player on each team, playing near the basket. On offense, the center tries to score on close shots and rebound. But on defense, the center tries to block opponents' shots and rebound their misses.

7 0

3 years ago

An atom has 5 protons, 6 neutrons, and 5 electrons. what is the atomic mass?

ozzi

Answer:

the atomic mass is 11

Explanation:

the atomic mass is basically how many protons and neutrons there are so for this all you have to do is some simple math:

5 + 6 = 11

and boom, ur atomic mass is equal to 11!

4 0

2 years ago