ME Project- Decomposer

1.) an object moves along the x axis, subject to the potential energy shown. The object has a mass of 1.1kg and starts at rest a

MrRa [10]

If I'm not wrong #1 should be C

4 0

4 years ago

Photo used to help with the question is below!! Please answer! Will mark BRAINLIEST!

Sergeu [11.5K]

This statement is True, Because the phenomenon is called a Mirage!

3 0

3 years ago

Read 2 more answers

A drag racer starts from rest and accelerates at 10 m/s squared for the

nika2105 [10]

Answer:

54 $\frac{m}{s}$

Explanation:

$v=u+at$

where $u=$ initial velocity

$v=0+(10\frac{m}{s^{2} } )$ × $(5.4s)$

$v=54\frac{m}{s}$

6 0

3 years ago

Why must we be careful when measuring current with a DMM?

Lemur [1.5K]

Answer:

DMM should be placed in the series combination with the circuit.

Explanation:

DMM is the digital multi meter. It can measure the voltage, current and resistance at a time.

While measuring the current with the DMM you must be ensure that the DMM should be connected with the circuit in series combination. So that it will give the resultant current accurately.

While measuring the voltage the observer should check the open probes.

4 0

3 years ago

A river flows due south with a speed of 2.0 m/s .You steer a motorboat across the river; your velocity relative to the water is

mihalych1998 [28]

Answer:

a) $v_m =\sqrt{v^2_x + v^2_y} = \sqrt{(2m/s)^2 +(4.8 m/s)^2}= 5.2 m/s$

b) $\theta= tan^{-1} \frac{v_y}{v_x} = tan^{-1} (\frac{2}{4.8})= 22.62$ degrees and on this case to the South of the East.

c) $t= \frac{w}{v_m}= \frac{600m}{4.8 m/s}= 125 s$

d) $Y = 2 m/s * 125 s = 250m$

So it would be 250 to the South

Explanation:

Part a

For this case the figure attached shows the illustration for the problem.

We know that $v_y = 2 m/s$ represent the velocity of the river to the south.

We have the velocity of the motorboard relative to the water and on this case is $V_x= 4.8 m/s$

And we want to find the velocity of the motord board relative to the Earth $v_m$

And we can find this velocity from the Pythagorean Theorem.

$v_m =\sqrt{v^2_x + v^2_y} = \sqrt{(2m/s)^2 +(4.8 m/s)^2}= 5.2 m/s$

Part b

We can find the direction with the following formula:

$\theta= tan^{-1} \frac{v_y}{v_x} = tan^{-1} (\frac{2}{4.8})= 22.62$ degrees and on this case to the South of the East.

Part c

For this case we can use the following definition

$D = Vt$

The distance would be D = w = 600 m and the velocity V = 4.8m/s and if we solve for t we got:

$t= \frac{w}{v_m}= \frac{600m}{4.8 m/s}= 125 s$

Part d

For this case we can use the same definition but now using the y compnent we have:

$Y = v_y t$

And replacing we got:

$Y = 2 m/s * 125 s = 250m$

So it would be 250 to the South

7 0

3 years ago