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3 years ago

HELP Giving all the points I can. Physics virtual lab report: circuit design Please help me.

Physics

2 answers:

mafiozo [28]3 years ago

5 0

Answer:text me I can help

Explanation:

Likurg_2 [28]3 years ago

3 0

Can you be more specific in the question. I doesn’t make sense

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(Q008) Magnetic anomalies are a. found only on the seafloor. b. measured with an instrument called an inclinometer. c. defined a

likoan [24]

Answer: D. places where the magnetic field strength is either greater or less than the expected strength.

Explanation:

Magnetic anomaly refers to the change in the magnitude of the magnetic field of the Eatth with respect to the location's expected value.

Magnetic anomalies are places where the magnetic field strength is either greater or less than the expected strength. Therefore, the correct option is D.

7 0

2 years ago

If a 5.5 kg object experiences 15 N of force for .15 seconds what is the speed change

Nady [450]

The speed change : Δv = 0.41 m/s

<h3>Further explanation</h3>

Given

mass = 5.5 kg

Force = 15 N

time = 0.15 s

Required

the speed change

Solution

Newton 2nd's law

Impulse and momentum

F = m.a

F = m . Δv/t

F.t = m.Δv

Input the value :

15 N x 0.15 s = 5.5 kg x Δv

Δv = 0.41 m/s

8 0

2 years ago

What is true of both gravity and magnetism?

stira [4]

Explanation:

I want to say option B - Both forces can act without objects touching.

5 0

2 years ago

For safety reasons, a worker’s eye travel time in a certain operation must be separated from the manual elements that follow. Th

iogann1982 [59]

Answer:

The answer is 12.67 TMU

Explanation:

Recall that,

worker’s eyes travel distance must be = 20 in.

The perpendicular distance from her eyes to the line of travel is =24 in

What is the MTM-1 normal time in TMUs that should be allowed for the eye travel element = ?

Now,

We solve for the given problem.

Eye travel is = 15.2 * T/D

=15.2 * 20 in/24 in

so,

= 12.67 TMU

Therefore, the MTM -1 of normal time that should be allowed for the eye travel element is = 12.67 TMU

7 0

3 years ago

You are driving at the speed of 27.7 m/s (61.9764 mph) when suddenly the car in front of you (previously traveling at the same s

densk [106]

Here when car in front of us applied brakes then it is slowing down due to frictional force on it

So here we can say that friction force on the car front of our car is given as

$F_f = \mu m g$

So the acceleration of car due to friction is given as

$F_{net} = - \mu mg$

$a = \frac{F_{net}}{m}$

$a = -\mu g$

now it is given that

$\mu = 0.868$

$g = 9.81 m/s^2$

so here we have

$a = -0.868 * 9.81$

$a = -8.52 m/s^2$

so the car will accelerate due to brakes by a = - 8.52 m/s^2

4 0

3 years ago