14.

QUESTION 3

Alisiya [41]

The force of frictions is opposed to relative motion.

The acceleration of the crate is approximately <u>2.937 m/s²</u>.

Reason:

The given parameters are;

The mass of the wood, m = 100 kg

The force which can move the wood, F = 588 N

Wood on wood static friction, $\mu_s$ = 0.5

Wood on wood kinetic friction, $\mu_k$ = 0.3

Solution;

The force of friction, $F_f$ , acting when the crate is moving is given as

follows;

$F_f$ = m × g × $\mu_k$

Where;

g = The acceleration due to gravity ≈ 9.81 m/s²

Therefore, we have;

$F_f$ = 100 × 9.81 × 0.3 = 294.3

The force of friction, $F_f$ = 294.3 N

The force with which the crate moves, F = 588 - 294.3 = 293.7

The force with which the crate moves, F = 293.7 N

Force = Mass, m × Acceleration, a

$a = \dfrac{F}{m}$

Therefore;

$a = \dfrac{293.7 \ N}{100 \ kg} = 2.937$

The acceleration of the crate, a ≈ <u>2.937 m/s²</u>.

Learn more about friction here:

brainly.com/question/94428

8 0

2 years ago

Bats are capable of navigating using the earth's field-a plus for an animal that may fly great distances from its roost at night

Nataliya [291]

Answer:

The correct magnitude of the coil's magnetic field= =50μT

Explanation :

The magnetic field takes place as a result of movement of charge I e current, can also occurs from magnetised material, magnetic field as a result of charge movement can be deducted using right hand grip rule.

At the equator magnetic field lines are parallel towards the earth's surface and the angle of inclination of the magnetic lines of force at the horizontal position is referred to as the angle of dip at the point.

As the current is produced then the varying magnetic field is opposed ,then there is induced current when the coil is positioned at varying magnetic field.

Given from the question,

angle below horizontal θ=60-degree

The Earth's magnetic field B=50μT

The horizontal magnetic field can be expressed in terms of the formula below;

BH=Bcos⁡θ

B(H) = earth's horizontal component of magnetic field

Ø is the angle between coil's field

B=the magnetic field in Tesla

Then,

BH=50μT×cos60∘

=50μT× 0.5

As the current is passed through the coil to produce a field , when combined with the earth's field, which creates a net field with the same strength and dip angle (60 degrees below horizontal) as the earth's field.

It can be deducted that B has the same magnitude and angle which makes the those vertical component to cancel each other since they are the same.

For the magnetic field to be pointed out at North direction, we can calculate the corrected magnetic field using the formula below

Bc=2BH

Bc=2×50μT× 0.5=50μT

The correct magnitude of the coil's magnetic field= =50μT

4 0

4 years ago

A 0.58 kg mass is moving horizontally with a speed of 6.0 m/s when it strikes a vertical wall. The mass rebounds with a speed of

Sunny_sXe [5.5K]

Answer:

$5.8\; {\rm kg\cdot m \cdot s^{-1}}$ .

Explanation:

If the mass of an object is $m$ and the velocity of that object is $v$ , the linear momentum of that object would be $m\, v$ .

Assume that the initial velocity of the mass is positive ( $6.0\; {\rm m\cdot s^{-1}}$ .) However, the direction of the velocity is reversed after the impact. Thus, the sign of the new velocity of the object would be negative- the opposite of that of the initial velocity. The new velocity would be $(-4.0\; {\rm m\cdot s^{-1}})$ .

Thus, the change in the velocity of the mass would be:

$\begin{aligned}& (\text{Change in Velocity}) \\ =\; & (\text{Final Velocity}) - (\text{Initial Velocity}) \\ =\; & (-4.0\; {\rm m\cdot s^{-1}}) - (6.0\; {\rm m\cdot s^{-1}}) \\ =\; & (-10\; {\rm m\cdot s^{-1})\end{aligned}$ .

The change in the linear momentum of the mass would be:

$\begin{aligned} & \text{change in momentum} \\ =\; & (\text{mass}) \times (\text{change in velocity}) \\ =\; & 0.58\; {\rm kg} \times (-10\; {\rm m\cdot s^{-1}}) \\ =\; & (-5.8\; {\rm kg \cdot m \cdot s^{-1}})\end{aligned}$ .

Thus, the magnitude of the change of the linear momentum would be $5.8\; {\rm kg \cdot m \cdot s^{-1}}$ .

7 0

3 years ago

What did Rutherford’s model of the atom include that Thomson’s model did not have?

Leona [35]

Rutherford tested Thomson'shypothesis by devising his "gold foil" experiment. Rutherford was forced to discard the Plum Pudding modeland reasoned that the only way the alpha particles could be deflected backwards was if most of the mass in an atom was concentrated in a nucleus.

6 0

3 years ago

Read 2 more answers

Please answer question B.

Flura [38]

Answer: 3.92

Explanation:

7 0

3 years ago