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3 years ago

A ball is thrown straight up with a launch of 3 m/s.

Physics

1 answer:

PtichkaEL [24]3 years ago

7 0

Explanation:

The ball is in free fall (gravity is the only force acting on the ball), so its acceleration is 9.8 m/s² down during the entire path.

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Neutrons can change the charge of an atom True False

IceJOKER [234]

Answer:

False. They can’t change the charge but they can change the mass

8 0

3 years ago

A speed skater is sliding across the ice at a speed of 19 m/s when she encounters a rough patch of ice that is 2.0 m wide. As sh

Stells [14]

Answer:

$-9.25 m/s^2$

Explanation:

The motion of the skater is a uniformly accelerated motion, so we can find the deceleration by using the following suvat equation:

$v^2-u^2=2as$

where

v is the final velocity

u is the initial velocity

a is the acceleration

s is the displacement

In this problem, we have:

u = 19 m/s is the initial velocity of the skater

v = 18 m/s is the final velocity of the skater

s = 2.0 m is the displacement

Solving for a, we can find the acceleration:

$a=\frac{v^2-u^2}{2s}=\frac{18^2-19^2}{2(2.0)}=-9.25 m/s^2$

And the negative sign tells us that the acceleration is in the opposite direction to the motion (so, it is a deceleration)

5 0

4 years ago

Two ions with masses of 5.29×10−27 kg move out of the slit of a mass spectrometer and into a region where the magnetic field is

riadik2000 [5.3K]

Answer:

0.132 m

Explanation:

$m$ = mass of the ion = 5.29 x 10⁻²⁷ kg

$q$ = magnitude of charge on singly charged ion = 1.6 x 10⁻¹⁹ C

$r$ = radius of circular path followed by singly charged ion

$v$ = speed of the ion = 1.13 x 10⁶ m/s

$B$ = magnitude of the magnetic field = 0.283 T

Radius of the circular path is given as

$r = \frac{mv}{qB}$

$r = \frac{(5.29\times 10^{-27})(1.13\times 10^{6})}{(1.6\times 10^{-19})(0.283)}$

$r$ = 0.132 m

6 0

3 years ago

The current in the wires of a circuit is 60.0 milliAmps. If the voltage impressed across the ends of the circuit were halved (i.

Ksivusya [100]

Answer:

30 miliAmps

Explanation:

Step 1:

Obtaining an expression to solve the question. This is illustrated below:

From ohm's law,

V = IR

Were:

V is the voltage.

I is the current.

R is resistance.

From the question given, we were told that the resistance is constant. Therefore the above equation can be written as shown below:

V = IR

V/I = constant

V1/I1 = V2/I2

V1 is initial voltage.

V2 final voltage.

I1 is initial current.

I2 final current.

Step 2:

Data obtained from the question. This include the following:

Initial voltage (V1) = V

Initial current (I1) = 60 miliAmps

Final voltage (V2) = one-half of the original voltage = 1/2V = V/2

Final current (I2) =..?

Step 3:

Determination of the new current. This can be obtained as follow:

V1/I1 = V2/I2

V/60 = (V/2) / I2

Cross multiply to express in linear form

V x I2 = V/2 x 60

V x I2 = V x 30

Divide both side by V

I2 = (V x 30)/V

I2 = 30mA.

Therefore, the new current is 30miliAmps

5 0

3 years ago

How would the speed of Earth's orbit around the sun change if Earth's distance from the sun

gayaneshka [121]

If Earth’s distance from the sun increased by 4 times then the speed of Earth’s orbit around the sun would decrease by a factor of 2.

Answer: Option 3

Explanation:

From Newton’s second law of motion, it is known that any kind of external force acting on an object will be equal to object’s mass and acceleration exerted on the object. So in this case, the gravitational force between Earth and Sun will be equal to the mass (M) and acceleration exerted on Earth.

$\text {Gravitational force}=\text {M} \times \text {Acceleration of Earth around the orbit}$

It is known that,

$\text {Acceleration in orbital motion}=\frac{\left(\text {velocity at which the Earth rotates)}^{2}\right.}{\text { Distance of Earth from the Sun}}$

Thus, substituting this, we get,

$\text {Gravitational force}=\frac{\text {M} \times\left(\text {velocity at which the Earth rotates)}^{2}\right.}{\text { Distance of Earth from the sun}}$

Also, we know,

$\text {Gravitational force}=\frac{G \times M \text { of sun } \times \text {M of Earth}}{(\text {Distance of Earth from the Sun})^{2}}$

Then, comparing both the equation, we get the orbital velocity as,

$\text { Orbital velocity } v=\sqrt{\frac{G \times M \text { of Sun }}{\text { Distance of Earth from the Sun }}}=\sqrt{\frac{G M}{r}}$

So, here G is gravitational constant, M is the mass of Sun and r is the distance of separation of Earth from Sun.

If the distance of Earth from Sun increases by 4 times so r’ will be 4r. Thus the new orbital velocity v’ will be

$v^{\prime}=\sqrt{\frac{G M}{r^{\prime}}}=\sqrt{\frac{G M}{4 r}}=\frac{1}{2} \sqrt{\frac{G M}{r}}$

So,

$v^{\prime}=\frac{1}{2} v$

Thus, the orbital speed will be decreased by a factor of 2 when the distance of Earth from the Sun increased by 4 times.

7 0

3 years ago