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2 years ago

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A man in a strength competition pulls an 18-wheel truck 3.10 m in 20.5 s. There is a cable that is attached to his body that exe

rts a horizontal force of 756 N. What is the power expended by the man?

1 answer:

larisa [96]2 years ago

4 0

Answer:

114.32195122 but Round your answer to three significant figures.) is 114

Explanation:

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If a rock is thrown upward on the planet mars with a velocity of 11 m/s, its height (in meters) after t seconds is given by h =

Butoxors [25]

(a) 3.56 m/s
(b) 11 - 3.72a
(c) t = 5.9 s
(d) -11 m/s
For most of these problems, you're being asked the velocity of the rock as a function of t, while you've been given the position as a function of t. So first calculate the first derivative of the position function using the power rule.
y = 11t - 1.86t^2
y' = 11 - 3.72t
Now that you have the first derivative, it will give you the velocity as a function of t.
(a) Velocity after 2 seconds.
y' = 11 - 3.72t
y' = 11 - 3.72*2 = 11 - 7.44 = 3.56
So the velocity is 3.56 m/s
(b) Velocity after a seconds.
y' = 11 - 3.72t
y' = 11 - 3.72a
So the answer is 11 - 3.72a
(c) Use the quadratic formula to find the zeros for the position function y = 11t-1.86t^2. Roots are t = 0 and t = 5.913978495. The t = 0 is for the moment the rock was thrown, so the answer is t = 5.9 seconds.
(d) Plug in the value of t calculated for (c) into the velocity function, so:
y' = 11 - 3.72a
y' = 11 - 3.72*5.913978495
y' = 11 - 22
y' = -11
So the velocity is -11 m/s which makes sense since the total energy of the rock will remain constant, so it's coming down at the same speed as it was going up.

3 0

2 years ago

The study of alternating electric current requires the solutions of equations of the form i equals Upper I Subscript max Baselin

KiRa [710]

Answer:

Explanation:

i = Imax sin2πft

given i = 180 , Imax = 200 , f = 50 , t = ?

Put the give values in the equation above

180 = 200 sin 2πft

sin 2πft = .9

sin2π x 50t = .9

sin 360 x 50 t = sin ( 360n + 64 )

360 x 50 t = 360n + 64

360 x 50 t = 64 , ( putting n = 0 for least value of t )

18000 t = 64

t = 3.55 ms .

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2 years ago

A jetboat is drifting with a speed of 5.0\,\dfrac{\text m}{\text s}5.0 s m 5, point, 0, start fraction, start text, m, end tex

love history [14]

The question is incomplete. Here is the entire question.

A jetboat is drifting with a speed of 5.0m/s when the driver turns on the motor. The motor runs for 6.0s causing a constant leftward acceleration of magnitude 4.0m/s². What is the displacement of the boat over the 6.0 seconds time interval?

Answer: Δx = - 42m

Explanation: The jetboat is moving with an acceleration during the time interval, so it is a <u>linear</u> <u>motion</u> <u>with</u> <u>constant</u> <u>acceleration</u>.

For this "type" of motion, displacement (Δx) can be determined by:

$\Delta x = v_{i}.t + \frac{a}{2}.t^{2}$

$v_{i}$ is the initial velocity

a is acceleration and can be positive or negative, according to the referential.

For Referential, let's assume rightward is positive.

Calculating displacement:

$\Delta x = 5(6) - \frac{4}{2}.6^{2}$

$\Delta x = 30 - 2.36$

$\Delta x$ = - 42

Displacement of the boat for t=6.0s interval is $\Delta x$ = - 42m, i.e., 42 m to the left.

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2 years ago

Irrigation channels that require regular flow monitoring are often equipped with electromagnetic flowmeters in which the magneti

san4es73 [151]

A) $1.36\cdot 10^{-4}T$

The magnetic field at the center of a coil of N turns is given by

$B=\frac{\mu_0 N I}{2R}$

where

I is the current in the coil

N is the number of turns

R is the radius of the coil

Here we have

I = 6.5 A is the current in the coil

N = 100 is the number of turns

$R=\frac{6.0 m}{2}=3.0 m$ is the radius of the coil

Substituting,

$B=\frac{(4\pi \cdot 10^{-7} H/m)(100)(6.5 A)}{2(3.0 m)}=1.36\cdot 10^{-4}T$

B) The force points north

The direction of the force on a positive ion in water can be found by using the right-hand rule. In fact, we have:

- Index finger: direction of motion of the ion --> towards east

- Middle finger: direction of magnetic field --> downward

- Thumb: direction of the force --> towards north

So, the force points north.

C) $3.26\cdot 10^{-23}N$

The magnitude of the magnetic force on a charged particle moving perpendicularly to the field is

$F=qvB$

where

q is the charge of the particle

v is the velocity

B is the magnitude of the magnetic field

In this case, we have

$q=+e=1.6\cdot 10^{-19} C$ is the charge

$v=1.5 m/s$ is the velocity

$B=1.36\cdot 10^{-4}T$ is the magnetic field strength

Substituting,

$F=(1.6\cdot 10^{-19} C)(1.5 m/s)(1.36\cdot 10^{-4}T)=3.26\cdot 10^{-23}N$

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2 years ago

Classify the sentences based on whether they describe temperature or thermal energy.

KiRa [710]

Thermal energy energy that has hot cold or warm
Temp.Energy is the out come of the thermal engery and affects Earth and space

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2 years ago