PLEASE HELP ME!!!

An object is thrown with an initial speed v near the surface of Earth. Assume that air resistance is negligible and the gravitat

IgorLugansk [536]

Answer:

E. downward and constant

Explanation:

Freefall is a special case of motion with constant acceleration because the acceleration due to gravity is always constant and downward. This is true even when an object is thrown upward or has zero velocity.

For example, when a ball is thrown up in the air, the ball's velocity is initially upward. Since gravity pulls the object toward the earth with a constant acceleration ggg, the magnitude of velocity decreases as the ball approaches maximum height. At the highest point in its trajectory, the ball has zero velocity, and the magnitude of velocity increases again as the ball falls back toward the earth.

3 0

3 years ago

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A metal wire is bent into a square and carries auniform current throughout it. The net magneticfield at the center of this squar

belka [17]

Answer:

The statement that the net magnetic field at the center of this square is zero is false.

The net magnetic field inside a conductor must be zero - This is a true statement

Explanation:

The net magnetic field at the center of this square is not equal to zero.

The net magnetic field at the center of this square is given by the equation below:

B = 2√2μ₀I/πₐ

Where a = the side of the loop, and I is the current.

Thus, the statement that the net magnetic field at the center of this square is zero is false.

The net magnetic field inside a conductor must be zero - This is a true statement because the total charge on the conductor must be equal to zero.

3 0

3 years ago

A circular loop of flexible iron wire has an initial circumference of 167 cm, but its circumference is decreasing at a constant

Kisachek [45]

Answer:

Part a)

$EMF = 5.6 \times 10^{-3} V$

Part b)

Since the radius is decreasing so induced current will increase the flux through the coil

So it would be clockwise in direction

Explanation:

As we know that magnetic flux linked with the coil is given as

$\phi = \pi r^2 B$

now the rate of change in flux is given as

$\frac{d\phi}{dt} = 2\pi r \frac{dr}{dt} B$

now we know that circumference is decreasing at rate of 15 cm/s

so here we know the length of circumference as

$C = 2\pi r$

So rate of change in circumference is

$\frac{dC}{dt} = 2\pi \frac{dr}{dt}$

$\frac{1}{2\pi}(15 cm) = \frac{dr}{dt}$

final length of circumference at t = 8 s

$C = 167 - (15)(8) = 47$

Part a)

Now the induced EMF is given as

$EMF = (2\pi r)(\frac{1}{2\pi})(0.15)(0.5)$

$EMF = (0.47)(\frac{1}{2\pi})(0.15)(0.5)$

$EMF = 5.6 \times 10^{-3} V$

Part b)

Since the radius is decreasing so induced current will increase the flux through the coil

So it would be clockwise in direction

5 0

3 years ago

A block of mass 21 kg slides up a frictionless plane inclined at 13◦ with the horizontal under the influence of a force of 210 N

svetlana [45]

Answer:

a= 8.74 m/s²

Explanation:

Given that

m = 21 kg

θ = 13◦

F= 210 N

g= 9.8 m/s²

The force due to gravity

F₁= m g sinθ

F₁=12 x 9.8 x sin 13◦

F₁=26.45 N

Lets take acceleration is a m/s²

Form Newtons second law

F - F₁ = m a

210 - 26.45 = 21 x a

a= 8.74 m/s²

3 0

3 years ago

A current balance is a device to measure magnetic forces. It is constructed from two parallel coils, each with an average radius

sveta [45]

Answer:

$I = 50.78 A$

Explanation:

Our Given Parameters include:

Average radius (r) for each parallel coil = 12.5 cm = 12.5 × 10⁻² m

The lower coil turns = 20 turns

The lower coil constant current = 1.30 A

The upper is suspended 0.314 cm above the lower coil

That indicates the distance = 0.314 cm = 0.314 × 10⁻² m

Current for the upper coil = ???(unknown)

Force (F) = 3.30 N

If we take each coil into cognizance as a long parallel straight wire; then the length of the lower coil can be calculated as:

$L__L}=N__L}(2 \pi r)$

where:

$N__L$ = number of turns of the lower coil = 20 turns

r = average radius in the lower coil = 12.5 × 10⁻² m

Substituting our values; we have:

$L__L}=20*(2 \pi (12.5*10^{-2}m)$

$L__L}=15.70m$

From our parameters above:

$I__L$ = constant current in the lower coil = 4.0 A

But the magnitude of the magnetic force (F) = 1 LB

Then the force on the lower coil in regard to the upper coil can be :

F = $I__L}L__L}[\frac{u__0I__upper}{2 \pi d}]$

Making the varied current in the upper coil the subject of the formula; we have:

$I_{upper}= \frac{2 \pi d F}{U_oI__L}L__L}$

where : $U__0}= 4 \pi *10^{-7}\frac{T.m}{A}$

Then:

$I_{upper}= \frac{2 \pi (0.314*10^{-2}(3.30)}{ (4 \pi *10^{-7}\frac{T.m}{A})(1.30A)(15.70m)}$

$I_{upper}= 2538.46 A$

≅ 2539 A

However; the current needed in the upper coil in each turn will be:

$I=\frac{I_{upper}}{50 turns}$

$I= \frac{2539}{50}$

$I = 50.78 A$

6 0

3 years ago