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ludmilkaskok [199]

4 years ago

12

An object moving north with an initial velocity of 14 m/s accelerates 5 m/s^2 for 20 seconds.what is the final velocity of the o

bject?
39 m/s
90 m/s
114 m/s
414m/s

2 answers:

bekas [8.4K]4 years ago

5 0

Answer:

Option C = 114 m/s

Explanation:

a=v-u/t

By substituting,

5=v-14/20

100=v-14

Thus, v=100+14

v= 114 m/s

Hope it helps :)

ELEN [110]4 years ago

5 0

The answer is 114 m/s good luck!!!

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9 m/s to a total stop. what is the objects acceleration

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We would need to know the time it took to slow to a stop.

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3 years ago

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8 0

3 years ago

A sled slides along a horizontal surface on which the coefficient of kinetic friction is 0.25. Its velocity at point A is 8.4 m/

Bad White [126]

Answer:

1.06 secs

Explanation:

Initial speed of sled, u = 8.4 m/s

Final speed of sled, v = 5.8 m/s

Coefficient of kinetic friction, μ = 0.25

Using the impulse momentum theory, we know that the impulse applied to the sled is equal to change in momentum of the sled:

FΔt = mv - mu

where m = mass of the object

Δt = time interval

F = force applied

The force applied on the sled is the frictional force, which is given as:

F = -μmg

where g = acceleration due to gravity

Therefore:

-μmgΔt = mv - mu

-μmgΔt = m(v - u)

-μgΔt = v - u

Making Δt subject of formula:

Δt = (v - u) / -μg

Δt = (5.8 - 8.4) / (-0.25 * 9.8)

Δt = -2.6/ -2.45

Δt = 1.06 secs

It took the sled 1.06 secs to travel from A to B.

7 0

3 years ago

Now find the electromotive force E2(t) induced across the entirety of solenoid 2 by the change in current in solenoid 1. Remembe

lions [1.4K]

Answer:

$E_{2} (t) = -\pi\mu_o} \rho^{2} n_{1}n_{2}L\frac{d }{dt}I_{1}(t)$

Explanation:

Consider two solenoids that are wound on a common cylinder as shown in fig. 1. Let the cylinder have radius 'ρ' and length 'L' .

No. of turns of solenoid 1 = n₁

No. of turns of solenoid 1 = n₂

Assume that length of solenoid is much longer than its radius, so that its field can be determined from Ampère's law throughout its entire length:

$\oint \overrightarrow {B}\overrightarrow {(r)}.\overrightarrow {dl}= \mu_{o}I$

We will consider the field that arises from solenoid 1, having n₁ turns per unit length. The magnetic field due to solenoid 1 passes through solenoid 2, which has n₂ turns per unit length.

Any change in magnetic flux from the field generated by solenoid 1 induces an EMF in solenoid 2 through Faraday's law of induction:

$\oint \overrightarrow {B}\overrightarrow {(r)}.\overrightarrow {dl}= -\frac{d}{dt} \phi _{M}(t)$

Consider B₁(t) magnetic feild generated in solenoid 1 due to current I₁(t)

Using:

$B_{1}(t) =\mu _{o} nI(t)\\$ --- (2)

Flux generated due to magnetic field B₁

$\phi _{1}(t) = \oint \overrightarrow {B_{1}}.dA\\$ ---(3)

area of solenoid = $A = \pi \rho^{2}$

substituting (2) in (3)

$\phi _{1}(t) = \mu_{o} \pi \rho^{2} n_{1}I_{1}(t)$ ----(4)

We have to find electromotive force E₂(t) induced across the entirety of solenoid 2 by the change in current in solenoid 1, i.e.

$E_{2} (t) = -n_{2}L\frac{d \phi_{1}}{dt}$ ---- (5)

substituting (4) in (5)

$E_{2} (t) = -n_{2}L\frac{d }{dt}(\mu_o} \pi \rho^{2} n_{1}I_{1}(t))\\E_{2} (t) = -\pi\mu_o} \rho^{2} n_{1}n_{2}L\frac{d }{dt}I_{1}(t)$

5 0

3 years ago

Each of 100 identical blocks sitting on a frictionless surface is connected to the next block by a massless string. The first bl

o-na [289]

Answer:

The tension in the string connecting block 50 to block 51 is 50 N.

Explanation:

Given that,

Number of block = 100

Force = 100 N

let m be the mass of each block.

We need to calculate the net force acting on the 100th block

Using second law of newton

$F=ma$

$100=100m\times a$

$ma=1\ N$

We need to calculate the tension in the string between blocks 99 and 100

Using formula of force

$F_{100-99}=ma$

$F_{100-99}=1$

We need to calculate the total number of masses attached to the string

Using formula for mass

$m'=(100-50)m$

$m'=50m$

We need to calculate the tension in the string connecting block 50 to block 51

Using formula of tension

$F_{50}=m'a$

Put the value into the formula

$F_{50}=50m\times a$

$F_{50}=50\times1$

$F_{50}=50\ N$

Hence, The tension in the string connecting block 50 to block 51 is 50 N.

8 0

3 years ago