You make a(n) _____ by fastening together usually perpendicular parts with the ends cut at an angle.

A construction crane, like the one shown, has a power output of 1,500 watts. If it takes 200 seconds to lift the roof to the top

Likurg_2 [28]

Answer:7.5j

Explanation:

P=w/s

4 0

3 years ago

The current in a wire varies with time according to the relationship 1=55A−(0.65A/s2)t2. (a) How many coulombs of charge pass a

mario62 [17]

Answer:

(A) Q = 321.1C (B) I = 42.8A

Explanation:

(a)Given I = 55A−(0.65A/s2)t²

I = dQ/dt

dQ = I×dt

To get an expression for Q we integrate with respect to t.

So Q = ∫I×dt =∫[55−(0.65)t²]dt

Q = [55t – 0.65/3×t³]

Q between t=0 and t= 7.5s

Q = [55×(7.5 – 0) – 0.65/3(7.5³– 0³)]

Q = 321.1C

(b) For a constant current I in the same time interval

I = Q/t = 321.1/7.5 = 42.8A.

3 0

3 years ago

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An object at rest cannot remain at rest unless which of the following holds?View Available Hint(s)An object at rest cannot remai

Leni [432]

Answer:

True The net force must be zero for the acceleration to be zero

Explanation:

In order to analyze the statements of this problem we propose your solution.

First let's look at Newton's first, which stable that every object is at rest or with constant speed unless something takes it out of this state (acceleration)

Now let's look at the second postulate, which says that force is related to the product of the mass of a body and its acceleration.

As a result of these two laws, for a body is a constant velocity the summation force on it must be zero.

Now we can analyze the statements given.

True The net force must be zero for the acceleration to be zero

False. If the force is different from zero, there is acceleration that changes the speeds

False. There may be forces, but the sum of them must be zero

False. If a force acts, the acceleration is different from zero and the speed changes

5 0

3 years ago

Sometimes i wish i wasnt alive....

pshichka [43]

Answer:

Swim, jogging, weight lifting are some activities to do. This will help the overall health of the heart and mind.

Explanation:

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3 years ago

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A satellite in outer space is moving at a constant velocity of 21.4 m/s in the y direction when one of its onboard thruster turn

kumpel [21]

Answer:

a) The magnitude of the satellite's velocity when the thruster turns off is approximately 24.177 meters per second.

b) The direction of the satellite's velocity when the thruster turns off is approximately 62.266º.

Explanation:

Statement is incomplete. The complete description is now described below:

A satellite in outer space is moving at a constant velocity of 21.4 m/s in the y direction when one of its onboard thruster turns on, causing an acceleration of 0.250 m/s2 in the x direction. The acceleration lasts for 45.0 s, at which point the thruster turns off.

(a) What is the magnitude of the satellite's velocity when the thruster turns off

(b) What is the direction of the satellite's velocity when the thruster turns off? Give your answer as an angle measured counterclockwise from the +x-axis. ° counterclockwise from the +x-axis

Let be x and y-directions orthogonal to each other and the satellite is accelerated uniformly from rest in the +x direction and moves at constant velocity in the +y direction. The velocity vector of the satellite ( $\vec{v}_{S}$ ), measured in meters per second, is:

$\vec{v}_{S} = (v_{o,x}+a_{x}\cdot t)\,\hat{i}+v_{y}\,\hat{j}$

Where:

$v_{o,x}$ - Initial velocity in +x direction, measured in meters per second.

$a_{x}$ - Acceleration in +x direction, measured in meter per square second.

$t$ - Time, measured in seconds.

$v_{y}$ - Velocity in +y direction, measured in meters per second.

If we know that $v_{o,x} = 0\,\frac{m}{s}$ , $a_{x} = 0.250\,\frac{m}{s^{2}}$ , $t = 45\,s$ and $v_{y} = 21.4\,\frac{m}{s}$ , the final velocity of the satellite is:

$\vec{v}_{S} = \left[0\,\frac{m}{s}+\left(0.250\,\frac{m}{s^{2}} \right)\cdot (45\,s) \right]\,\hat{i}+\left(21.4\,\frac{m}{s} \right)\,\hat{j}$

$\vec{v_{S}} = 11.25\,\hat{i}+21.4\,\hat{j}\,\,\left[\frac{m}{s} \right]$

a) The magnitud of the satellite's velocity can be found by the resource of the Pythagorean Theorem:

$\|\vec {v}_{S}\| = \sqrt{\left(11.25\,\frac{m}{s} \right)^{2}+\left(21.4\,\frac{m}{s} \right)^{2}}$

$\|\vec{v}_{S}\| \approx 24.177\,\frac{m}{s}$

The magnitude of the satellite's velocity when the thruster turns off is approximately 24.177 meters per second.

b) The direction of the satellite's velocity when the thruster turns off is determined with the help of trigonometric functions:

$\tan \alpha = \frac{v_{y}}{v_{x}} = \frac{21.4\,\frac{m}{s} }{11.25\,\frac{m}{s} }$

$\tan \alpha = 1.902$

$\alpha = \tan^{-1}1.902$

$\alpha \approx 62.266^{\circ}$

The direction of the satellite's velocity when the thruster turns off is approximately 62.266º.

4 0

3 years ago