Will give brainliest!!

3. A man hits a 0.2 kg golf ball with a force of 4

pogonyaev

Answer:

Force(f)= mass x acceleration

Acceleration (a) is the rate of change in velocity.

F=4N

M=0.2kg

a=F/M

a=4/0.2

a=20m/s^2

Explanation:

8 0

2 years ago

If a 5000-kg is moving at a speed of 43 m/s, what is its momentum?

Eduardwww [97]

Answer:

215000kgm/s

Explanation:

Given parameters:

Mass of the moving body = 5000kg

Velocity = 43m/s

Unknown:

Momentum = ?

Solution:

The momentum of a body is the amount of motion a body possess.

It is mathematically expressed as:

Momentum = mass x velocity

Now:

Momentum = 5000 x 43 = 215000kgm/s

4 0

2 years ago

A transformer consists of a 500 turn primary coil and a 2000-turnsecondary coil. If the current in the secondary is 3.0A, what i

Neporo4naja [7]

Answer:

The correct solution will be "12.0 A".

Explanation:

The given values are:

$N_p= 500 \ turn$

$N_s= 200 \ turn$

$I_s= 3.0 \ A$

By using the transformer formula, we get

⇒ $\frac{N_p}{N_s} =\frac{I_s}{I_p}$

⇒ $I_p = I_s\times \frac{N_s}{N_p}$

On substituting the given values, we get

⇒ $=3.0 \ A\times \frac{2000}{500}$

⇒ $=12.0 \ A$

8 0

3 years ago

A student conducts an experiment in which a cart is pulled by a variable applied force during a 2 s time interval. In trial 1, t

fredd [130]

Answer:

The answer is "Including all three studies of 0s to 2s, that shift in momentum is equal".

Explanation:

Its shift in momentum doesn't really depend on the magnitude of its cars since the forces or time are similar throughout all vehicles.

Let's look at the speed of the car

$F = m a\\\\a =\frac{F}{m}$

We use movies and find lips

$\to v = v_0 + a t\\\\\to v = v_0 + (\frac{F}{m}) t$

The moment is defined by

$\to p = m v$

The moment change

$\Delta p = m v - m v_0$

Let's replace the speeds in this equation

$\Delta p = m (v_0 + \frac{F}{m t}) - m v_0\\\\\Delta p = m v_0 + F t - m v_0\\\\\Delta p = F t$

They see that shift is not directly proportional to the mass of cars since the force and time were the same across all cars.

5 0

2 years ago

xConsider the following reduction potentials: Cu2+ + 2e– Cu E° = 0.339 V Pb2+ + 2e– Pb E° = –0.130 V For a galvanic cell employi

slega [8]

Answer:

Approximately $\rm 90\; kJ$ .

Explanation:

Cathode is where reduction takes place and anode is where oxidation takes place. The potential of a electrochemical reaction ( $E^{\circ}(\text{cell})$ ) is equal to

$E^{\circ}(\text{cell}) = E^{\circ}(\text{cathode}) - E^{\circ}(\text{anode})$ .

There are two half-reactions in this question. $\rm Cu^{2+} + 2\,e^{-} \rightleftharpoons Cu$ and $\rm Pb^{2+} + 2\,e^{-} \rightleftharpoons Pb$ . Either could be the cathode (while the other acts as the anode.) However, for the reaction to be spontaneous, the value of $E^{\circ}(\text{cell})$ should be positive.

In this case, $E^{\circ}(\text{cell})$ is positive only if $\rm Cu^{2+} + 2\,e^{-} \rightleftharpoons Cu$ is the reaction takes place at the cathode. The net reaction would be

$\rm Cu^{2+} + Pb \to Cu + Pb^{2+}$ .

Its cell potential would be equal to $0.339 - (-0.130) = \rm 0.469\; V$ .

The maximum amount of electrical energy possible (under standard conditions) is equal to the free energy of this reaction:

$\Delta G^{\circ} = n \cdot F \cdot E^{\circ} (\text{cell})$ ,

where

$n$ is the number moles of electrons transferred for each mole of the reaction. In this case the value of $n$ is $2$ as in the half-reactions.
$F$ is Faraday's Constant (approximately $96485.33212\; \rm C \cdot mol^{-1}$ .)

$\begin{aligned}\Delta G^{\circ} &= n \cdot F \cdot E^{\circ} (\text{cell})\cr &= 2\times 96485.33212 \times (0.339 - (-0.130)) \cr &\approx 9.0 \times 10^{4} \; \rm J \cr &= 90\; \rm kJ\end{aligned}$ .

5 0

2 years ago