Ask question

Ask question

All categories

3 years ago

10

An object is thrown upwards with a velocity of 3 meters per second, such that the only force acting on it is gravity (accelerati

on is -9.8). What will the object's velocity be after 2 seconds?
A.) -16.6 meters per second (down)
B.) 16.6 meters per second (up)
C.) -22.6 meters per second (down)
D.) 22.6 meters per second (up)
E.) 3 meters per second (up)

2 answers:

3241004551 [841]3 years ago

3 0

Answer:

Correct answer: A.) V = - 16.6 m/s down

Explanation:

Given:

V₀ = 3 m/s initial velocity

t = 2 seconds

g = 9.8 m/s²

V(t) = V(2) = ?

The movement described is a vertical upward shot

For velocity at any time is valid the next formula

V = V₀ - g · t

V = 3 - (9.8 · 2) = 3 - 19.6 = - 16.6 m/s down

Under condition that it has a enough drop height with respect to the ejection point.

God is with you!!!

NNADVOKAT [17]3 years ago

3 0

The answer is A. DOWN

You might be interested in

Two cars start from rest at a red stop light. When the light turns green, both cars accelerate forward. The blue car accelerates

saul85 [17]

First, create an illustration of the motion of the two cars as shown in the attached picture. The essential equations used is

For constant acceleration:
a = v,final - v,initial /t

The solutions is as follows:

a = v,final - v,initial /t
3.8 = (v - 0)/2.8 s
v = 10.64 m/s
After 2.8 seconds, the speed of the blue car is 10.64 m/s.

4 0

3 years ago

46. Nuclear generating plants use the potential

nydimaria [60]

Answer:

nuclear battery to generate energy

4 0

3 years ago

An electron is moving at 7.4x105 m/s perpendicular to a magnetic field. It experiences a force of 2.0x10–13 N. What is the magne

alisha [4.7K]

Answer:

1.69 T

Explanation:

Applying,

F = BvqsinФ.................. Equation 1

Where F = Force, B = magnetic field, v = velocity, q = charge on an electron, Ф = angle between the electron and the field.

make B the subject of the equation,

B = F/(vqsinФ)............. Equation 2

From the question,

Given: F = 2.0×10⁻¹³ N, v = 7.4×10⁵ m/s, Ф = 90°

Constant: q = 1.60×10⁻¹⁹ C

Substitute into equation 2

B = 2.0×10⁻¹³/(7.4×10⁵×1.60×10⁻¹⁹×sin90°)

B = 0.169×10

B = 1.69 T

4 0

3 years ago

A multimeter in an RL circuit records an rms current of 0.600 A and a 50.0-Hz rms generator voltage of 110 V. A wattmeter shows

jeka57 [31]

Answer:

(a) The impedance in the circuit is $Z=183.33\Omega$ .

(b)The resistance is $R=38.89\Omega$ .

(c) The inuctance is 0.57 H.

Explanation:

(a)

The expression for the impedance is as follows:

$Z=\frac{V_rms}{I_rms}$

Here, $V_rms$ is the rms voltage and $I_rms$ is the rms current.

Put $V_rms=110 V$ and $I_rms=0.600 A$ .

$Z=\frac{110}{0.600}$

$Z=183.33\Omega$

Therefore, the impedance in the circuit is $Z=183.33\Omega$ .

(b)

The expression for the average power is as follows;

$P_{a}=I_{rms}^{2}R$

Here, $P_{a}$ is the average power and R is the resistance.

Calculate the resistance by rearranging the above expression.

$R=\frac{P_{a}}{I_{rms}^{2}}$

Put $P_{a}=14W$ and

$R=\frac{14}{{0.600}^{2}}$

$R=38.89\Omega$

Therefore, the resistance is $R=38.89\Omega$ .

(c)

The expression for the impedance is as follows;

$Z^{2}=R^{2}+X_{L}^{2}$

Here, $X_{L}$ is the inductive reactance.

Put $Z=183.33\Omega$ and $R=38.89\Omega$ .

$(183.33)^{2}=(38.89)^{2}+X_{L}^{2}$

$X_{L}=179.16\Omega$

The expression for the inductive reactance in terms of frequency is as follows;

$X_{L}=2\pi fL$

Here, L is the inductance.

Calculate the inductance by rearranging the above expression.

$L=\frac{X_{L}}{2\pi f}$

Put $X_{L}=179.16\Omega$ and f=50Hz.

$L=\frac{179.16}{2\pi (50)}$

L=0.57 H

Therefore, the inuctance is 0.57 H.

4 0

3 years ago

How does a rotating coil inside a magnetic field generate electricity?

Goryan [66]

Answer:

When an electrical current passes through a wire, a magnetic field is generated around it. Likewise, if the magnetic field around a wire is changed ( for example by rotating a coil inside a stationary manger), electricity will move through the wire.

4 0

3 years ago