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3 years ago

PLEASE HELP!!! GIVING BRAINLIEST!! ill also answer questions that you have posted if you answer this correctly!!!! (60pts)

Physics

1 answer:

mr Goodwill [35]3 years ago

3 0

The density of an object determines whether it will float or sink in another substance. An object will float if it is less dense than the liquid it is placed in. An object will sink if it is more dense than the liquid it is placed in.
So since the boat has a lower density than the water, it will float.
So the answer is choice B

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At each corner of a square of side l there are point charges of magnitude Q, 2Q, 3Q, and 4Q.What is the magnitude and direction

lbvjy [14]

Answer:

$F_T=6k\frac{Q^2}{L}\hat{i}+10k\frac{Q^2}{L}\hat{j}=2k\frac{Q^2}{L}[3\hat{i}+5\hat{j}]$

$|F_T|=2\sqrt{34}k\frac{Q^2}{L}$

$\theta=tan^{-1}(\frac{5}{3})=59.03\°$

Explanation:

I attached an image below with the scheme of the system:

The total force on the charge 2Q is the sum of the contribution of the forces between 2Q and the other charges:

$F_T=F_Q+F_{3Q}+F_{4Q}\\\\F_T=k\frac{(Q)(2Q)}{R_1}\hat{i}+k\frac{(3Q)(2Q)}{R_2}\hat{j}+k\frac{(4Q)(2Q)}{R_3}[cos\theta \hat{i}+sin\theta \hat{j}]$

the distances R1, R2 and R3, for a square arrangement is:

R1 = L

R2 = L

R3 = (√2)L

θ = 45°

$F_T=k\frac{2Q^2}{L}\hat{i}+k\frac{6Q^2}{L}\hat{j}+k\frac{8Q^2}{\sqrt{2}L}[cos(45\°)\hat{i}+sin(45\°)\hat{j}]\\\\F_T=k\frac{2Q^2}{L}\hat{i}+k\frac{6Q^2}{L}\hat{j}+k\frac{8Q^2}{\sqrt{2}L}[\frac{\sqrt{2}}{2}\hat{i}+\frac{\sqrt{2}}{2}\hat{j}]\\\\F_T=6k\frac{Q^2}{L}\hat{i}+10k\frac{Q^2}{L}\hat{j}=2k\frac{Q^2}{L}[3\hat{i}+5\hat{j}]$

and the magnitude is:

$|F_T|=2k\frac{Q^2}{L}\sqrt{3^2+5^2}=2\sqrt{34}k\frac{Q^2}{L}$

the direction is:

$\theta=tan^{-1}(\frac{5}{3})=59.03\°$

4 0

3 years ago

What is potential energy called that is associated with objects that can be stretched

Umnica [9.8K]

Elastic potential energy

5 0

3 years ago

Assuming a 8 kilogram bowling ball moving at 2 m/s bounces off a spring at the same speed that had before bouncing what is the a

Naya [18.7K]

a) 32 kg m/s

Assuming the spring is initially at rest, the total momentum of the system before the collision is given only by the momentum of the bowling ball:

$p_i = m u = (8 kg)(2 m/s)=16 kg m/s$

The ball bounces off at the same speed had before, but the new velocity has a negative sign (since the direction is opposite to the initial direction). So, the new momentum of the ball is:

$p_{fB}=m v_b =(8 kg)(-2 m/s)=-16 kg m/s$

The final momentum after the collision is the sum of the momenta of the ball and off the spring:

$p_f = p_{fB}+p_{fS}$

where $p_{fS}$ is the momentum of the spring. For the conservation of momentum,

$p_i = p_f\\p_i = p_{fB}+p_{fS}\\p_{fS}=p_i -p_{fB}=16 kg m/s -(-16 kg m/s)=32 kg m/s$

b) -32 kg m/s

The change in momentum of bowling ball is given by the difference between its final momentum and initial momentum:

$\Delta p=p_{fb}-p_i=-16 kg m/s - 16 kg m/s=-32 kg m/s$

c) 64 N

The change in momentum is equal to the product between the average force and the time of the interaction:

$\Delta p=F \Delta t$

Since we know $\Delta t=0.5 s$ , we can find the magnitude of the force:

$F=\frac{\Delta p}{\Delta t}=\frac{-32 kg m/s}{0.5 s}=-64 N$

The negative sign simply means that the direction of the force is opposite to the initial direction of the ball.

d) The force calculated in the previous step (64 N) is larger than the force of 32 N.

5 0

3 years ago

a horse moves a sleigh 1.00 kilometers by applying a horizontal 2000 Newton force on its harness for 45 minutes. what is the pow

tatyana61 [14]

-- If 2,000 newtons of force were applied through a distance of 1,000 meters,
then 2,000,000 newton-meters = 2,000,000 joules of work were done.

-- 45 minutes = (45 x 60) = 2,700 seconds

-- Power = (work) / (time) = (2,000,000 j) / (2,700 s) = <u>740.74 watts</u>

Interestingly, that's almost exactly 1 horsepower. (0.99295... of 746 watts)

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3 years ago

Would a pair of scissors with really long blades and a short handle be practical? Why or why not?

Lana71 [14]

Yes, for balance.hope this helped.

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3 years ago