1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Deffense [45]
3 years ago
9

Technician A says that the starter motor used to crank diesel engines can draw up to 400 amps of current. Technician B says that

high resistance on the insulated side of a starter motor circuit would cause higher than specified starter motor current draw. Who is correct?a. Technician A
b. Technician B
c. Both Technician A and Technician B
d. Neither Technician A nor Technician B
Physics
1 answer:
Aleks04 [339]3 years ago
8 0

Answer: Option A : Technician A

Explanation:

The statement/observation, "that the starter motor used to crank diesel engines can draw up to 400 amps of current" made by Technician A is correct.

A diesel engine uses up to 400+ Amperes of electricity to start up a diesel engine in the ignition chamber of motor engine.

You might be interested in
An object ends up at a final position of x=-55.25 meters after a displacement of -189.34 meters after a displacement of -189.34
Travka [436]

The initial position of the object was found to be 134.09 m.

<u>Explanation:</u>

As displacement is the measure of difference between the final and initial points. In other words, we can say that displacement can be termed as the change in the position of the object irrespective of the path followed by the object to change the path. So

Displacement = Final position - Initial position.

As the final position is stated as -55.25 meters and the displacement is also stated as -189.34 meters. So the initial position will be

Initial position of the object = Final position-Displacement

Initial position = -55.25 m - (-189.34 m) = -55.25 m + 189.34 m = 134.09 m.

Thus, the initial position for the object having a displacement of -189.34 m is determined as 134.09 m.

4 0
3 years ago
A dog exerts a force of 30N to move a wagon 2m in 5s. What is the power of the dog
Hunter-Best [27]

Explanation:

power=f×v. recall= distances/ time

= f× d/t

= 30 × 2/5

=12watt

6 0
3 years ago
What is the approximate value of the gravitational force between a 73 kg astronaut and a 7.1×104 kg spacecraft when they're 89 m
Luden [163]

Answer:

F = 4.3671 * 10^{-8}\ Newtons

Explanation:

The gravitational force between two corpses is given by the following equation:

F = GMm/d^2

Where F is the force, G is the gravitational constant

(G = 6.67408*10^{-11}\ m^3kg^{-1}s^{-2}), M and m are the masses of the corpses and d is the distance between them.

So we have that:

F = 6.67408*10^{-11} * 7.1*10^4 * 73/89^2

F = 4.3671 * 10^{-8}\ Newtons

5 0
3 years ago
A wire loop of radius 0.37 m lies so that an external magnetic field of magnitude 0.35 T is perpendicular to the loop. The field
Galina-37 [17]

Answer:

168.57 mV

Explanation:

Initial magnetic flux = BA , B magnetic field and A is area of loop

= .35 x 3.14 x .37²

= .15 Weber

Final magnetic flux

= - .2 x 3.14 x .37²

= -  .086 Weber

change in flux

.15 +  .086

= .236 Weber

rate of change of flux

= .236 / 1.4

= .16857 V

= 168.57 mV

5 0
3 years ago
In February 1955, a paratrooper fell 370 m from an airplane without being able to open his chute but happened to land in snow, s
nevsk [136]

a) 0.94 m

The work done by the snow to decelerate the paratrooper is equal to the change in kinetic energy of the man:

W=\Delta K\\-F d = \frac{1}{2}mv^2 - \frac{1}{2}mu^2

where:

F=1.1 \cdot 10^5 N is the force applied by the snow

d is the displacement of the man in the snow, so it is the depth of the snow that stopped him

m = 68 kg is the man's mass

v = 0 is the final speed of the man

u = 55 m/s is the initial speed of the man (when it touches the ground)

and where the negative sign in the work is due to the fact that the force exerted by the snow on the man (upward) is opposite to the displacement of the man (downward)

Solving the equation for d, we find:

d=\frac{1}{2F}mu^2 = \frac{(68 kg)(55 m/s)^2}{2(1.1\cdot 10^5 N)}=0.94 m

b) -3740 kg m/s

The magnitude of the impulse exerted by the snow on the man is equal to the variation of momentum of the man:

I=\Delta p = m \Delta v

where

m = 68 kg is the mass of the man

\Delta v = 0-55 m/s = -55 m/s is the change in velocity of the man

Substituting,

I=(68 kg)(-55 m/s)=-3740 kg m/s

7 0
3 years ago
Other questions:
  • What name is given to a process where a substance melts under pressure?​
    7·1 answer
  • A 69 kg driver gets into an empty taptap to start the day's work. The springs compress 2×10−2 m . What is the effective spring c
    10·1 answer
  • If you measured all the energy related to motion and all the stored energy in the particles of a substance, which would you be m
    13·1 answer
  • How are moving pulleys different from fixed-position pulleys? (Points : 3) Moving pulleys have more force than fixed-position pu
    5·2 answers
  • An isotope of an element will be radioactive if it has an unstable ratio of____and_____
    15·1 answer
  • Please help!!! These questions are about specific heat capacity.
    13·1 answer
  • Two asteroids are 100,000 m apart. One has a mass of 3.5 x 106 kg. If the
    14·2 answers
  • The work done in lifting a brick of mass 2kg through a height of 5m above the ground will be
    8·1 answer
  • a toy car travels from A to B at constant speed 30km/hr and without stopping at B return A at constant speed v. if the average s
    8·1 answer
  • A 20 kg sled is pulled up a 10m tall hill. What work is done against gravity?
    9·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!