Please help!!!!!!!!!!

Please answer the questions here in the attachment I attached below. They are all multiple choice.

Radda [10]

Answer:

abcbc

Explanation:

8 0

4 years ago

The law that states the volume of gas is directly proportional to its temperature in kelvins if the pressure and the number of p

lorasvet [3.4K]

The answer is d) Charles's law.

7 0

3 years ago

Two tugboats pull a barge across the harbor. One boat exerts a force of 7.5 x 10^4N north, while the second boat exerts a force

Sonbull [250]

The barge moves at an angle <u>47.3°North of West.</u>

The tug boats exert forces in different directions on the barge and the barge moves in the direction of the resultant force.

Draw a diagram representing the forces, as shown in the figure attached. The tug boat 1 exerts a force F₁ along North while the tug boat 2 exerts a force F₂ at an angle of 15° North of west.

Resolve the forces along the North and the West.

The force acting along the North is given by,

$F_N=F_1+F_2sin15\\ =(7.5*10^4N)+(9.5*10^4 N*sin15)\\ =9.96*10^4N$

The force along the West is given by,

$F_W=F_2cos15=(9.5*10^4N)cos15\\ =9.18*10^4N$

The resultant force <em>F </em> makes an angle θ North of West.

Therefore, from the diagram,

$tan\theta =\frac{F_N}{F_W} \\ =\frac{9.96*10^4N}{9.18*10^4N} \\ =1.0849$

Hence,

$\theta =tan^-^1(1.00849)\\ =47.3^o$

Thus, the barge moves in a direction 47.3° North of West.

7 0

3 years ago

You drop a ball from a window on an upper floor of a building and it is caught by a friend on the ground when the ball is moving

Marysya12 [62]

Answer:1.44 s,10.17 m

Explanation:

Given

two balls are separated by a distance of 41.1 m

One person drops the ball from a height of 41.1 and another launches a ball with velocity of 41.1 m exactly at the same time.

Let the ball launches from ground travels a distance of x m in t sec

For Person on window

$41.1-x=ut+\frac{1}{2}gt^2$

$41.1-x=0+\frac{1}{2}\times 9.81\times t^2--------1$

For person at ground

$x=v_ft-\frac{1}{2}gt^2---------2$

add (1) & (2)

$41.1=v_ft-\frac{1}{2}gt^2+\frac{1}{2}gt^2$

$41.1=v_ft$

and $v_f$ is given by

$v_f=\sqrt{2\times 9.81\times 41.1}=28.39 m/s$

$t=\frac{41.1}{28.39}=1.44 s$

Substitute value of t in equation 1

$41.1-x=0+\frac{1}{2}\times 9.81\times 1.44^2$

41.1-x=10.171 m

Thus the two ball meet at distance of 10.17 m below the window.

8 0

3 years ago

How does jumping into a pool change the amount of gravity acting on you?

WINSTONCH [101]

Answer:

Because of the buoyant force

Explanation:

Jumping into a pool change the amount of apparent gravity acting on a person.

Normally, for a person in free fall, the weight of the person is given by:

$F=mg$

where m is the mass of the person and g=9.8 m/s^ is the acceleration due to gravity.

When a person is in the water, there is a buoyant force pushing the person upward. The magnitude of the buoyant force is

$B=\rho_w V g$

where

$\rho_w$ is the density of the water

V is the volume of displaced fluid

g is the acceleration due to gravity

So the net force acting on the person is

$F= mg - \rho_w V g$ (1)

Since V corresponds to the volume of the person, we can rewrite it as

$V=\frac{m}{\rho_p}$

where $\rho_p$ is the density of the person. Substituting into eq.(1),

$F=mg- m \frac{\rho_w}{\rho_p} g = mg - mg'$ (2)

where we called

$g'=\frac{\rho_w}{\rho_p} g$

So we can further rewrite (2) as

$F=m(g-g')$

so we see that the gravity acting on the person, g, has been modified into (g-g') due to the presence of the buoyant force.

4 0

3 years ago