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2 years ago

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You have managed to get a vacation on the Moon where the gravitational field strength is 1.6 N/kg. Gleefully, you kick a ball of

mass 1.25 kg with an initial velocity of < 12, 12, 0> m/s. There is a goal post you can use as an origin marker where you kicked the ball. What is the initial momentum of the ball?

1 answer:

sattari [20]2 years ago

4 0

The initial momentum of the ball is 15 Kgms-1.

<h3>What is momentum?</h3>

In Physics, the term momentum is the product of mass and velocity. Momentum is a vector quantity. This implies that momentum possess both magnitude and direction.

Hence, we often write; p = mv

Where:

p = momentum

m = mass

v = velocity

The initial momentum therefore is;

p = 1.25 kg × 12 m/s = 15 Kgms-1

Learn more about momentum: brainly.com/question/904448

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A weightlifter exerts an upward force on a 1000-N barbell and holds it at a height of 1 meter for 2 seconds. Approximately how m

Alex73 [517]

Amount of work done is zero and so power = 0 watts.

<u>Explanation:</u>

Power is the rate at which work is done, or W divided by delta t. Since the barbell is not moving, the weightlifter is not doing work on the barbell.Therefore, if the work done is zero, then the power is also zero.It may seem unusual that the data given in question is versatile i.e. A weightlifter exerts an upward force on a 1000-N barbell and holds it at a height of 1 meter for 2 seconds. But, still the answer is zero watts , this was a tricky question although conceptual basis of question was good! Power is dependent on amount of work done which is further related to displacement and here the net displacement is zero ! Hence, amount of work done is zero and so power = 0 watts.

6 0

3 years ago

As the concentration of a solute in a solution increases, the freezing point of the solution ________ and the vapor pressure of

kykrilka [37]

Answer:

As the concentration of a solute in a solution increases, the freezing point of the solution <u><em>decrease </em></u>and the vapor pressure of the solution <em><u>decrease </u></em>.

Explanation:

Depression in freezing point :

$\Delta T_f=K_f\times m$

where,

$\Delta T_f$ =depression in freezing point =

$K_f$ = freezing point constant

m = molality ( moles per kg of solvent) of the solution

As we can see that from the formula that higher the molality of the solution is directly proportionate to the depression in freezing point which means that:

If molality of the solution in high the depression in freezing point of the solution will be more.
If molality of the solution in low the depression in freezing point of teh solution will be lower .

Relative lowering in vapor pressure of the solution is given by :

$\frac{p_o-p_s}{p_o}=\chi_{solute}$

$p_o$ = Vapor pressure of pure solvent

$p_s$ = Vapor pressure of solution

$\chi_{solute}$ = Mole fraction of solute

$p_s\propto \frac{1}{\chi_{solute}}$

Vapor pressure of the solution is inversely proportional to the mole fraction of solute.

Higher the concentration of solute more will the be solute's mole fraction and decrease in vapor pressure of the solution will be observed.
lower the concentration of solute more will the be solute's mole fraction and increase in vapor pressure of the solution will be observed.

8 0

2 years ago

Influenced by the gravitational pull of a distant star, the velocity of an asteroid changes from from +19.3 km/s to −18.8 km/s o

muminat

As per above given data

initial velocity = 19.3 km/s

final velocity = - 18.8 km/s

now in order to find the change in velocity

$\Delta v = v_f - v_i$

$\Delta v = -18.8 - 19.3$

$\Delta v = -38.1 km/s$

$\Delta v = -3.81 * 10^4 m/s$

Part b)

Now we need to find acceleration

acceleration is given by formula

$a = \frac{\Delta v}{\Delta t}$

given that

$\Delta v =- 3.81 * 10^4 m/s$

$\Delta t = 2.07 years = 6.53 * 10^7 s$

now the acceleration is given as

$a = \frac{-3.81 * 10^4}{6.53 * 10^7}$

$a = - 5.84 * 10^{-4}m/s^2$

so above is the acceleration

4 0

3 years ago

An electron has a mass of 9.1x10-31 kg. What is

zloy xaker [14]

Answer:

3.185×10^-29 kgm/s

Explanation:

Momentum(p)=mass×velocity

=9.1×10^-31×3.5×10

=3.185×10^-29 kgm/s

4 0

3 years ago

A proton is placed in an electric field of intensity 700 N/C. What are the magnitude and direction of the acceleration of this p

olya-2409 [2.1K]

Answer:

Acceleration of proton will be $a=0.67\times 10^{11}m/sec^2$

Explanation:

We have given a proton is placed in an electric field of intensity of 700 N/C

So electric field E = 700 N/C

Mass of proton $m=1.67\times 10^{-27}kg$

Charge on proton $e=1.6\times 10^{-19}C$

So electric force on the proton $F=qE=1.6\times 10^{-19}\times 700=1.120\times 10^{-16}N$

This force will be equal to force due to acceleration of the proton

According to newton's law force is given by F = ma

So $1.67\times 10^{-27}\times a=1.120\times 10^{-16}$

$a=0.67\times 10^{11}m/sec^2$

So acceleration of proton will be $a=0.67\times 10^{11}m/sec^2$

6 0

3 years ago