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2 years ago

PLEASE HELP ASAP!!!!!

1 answer:

8 0

While falling, both the sheet of paper and the paper ball experience air resistance. But the surface area of the sheet is much more than that of the spherical ball. And air resistance varies directly with surface area. Hence the sheet experiences more air resistance than the ball and it falls more slowly than the paper ball.

Hope that helps!

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A trick shot archer shoots an arrow with a velocity of 30.0 m/s at an angle of 20.0 degrees with respect to the horizontal. An a

svlad2 [7]

Answer:

$u'=10.259\ m.s^{-1}$ is the initial velocity of tossing the apple.

the apple should be tossed after $\Delta t=0.0173\ s$

Explanation:

Given:

velocity of arrow in projectile, $v=30\ m.s^{-1}$

angle of projectile from the horizontal, $\theta=20^{\circ}$

distance of the point of tossing up of an apple, $d=30\ m$

Now the horizontal component of velocity:

$v_x=v\ cos\ \theta$

$v_x=30\times cos\ 20^{\circ}$

$v_x=28.191\ m.s^{-1}$

The vertical component of the velocity:

$v_y=v.sin\ \theta$

$v_y=30\times sin\ 20^{\circ}$

$v_y=10.261\ m.s^{-1}$

Time taken by the projectile to travel the distance of 30 m:

$t=\frac{d}{v_x}$

$t=\frac{30}{28.191}$

$t=1.0642\ s$

Vertical position of the projectile at this time:

$h=v_y.t-\frac{1}{2}g.t^2$

$h=10.261\times 1.0642-\frac{1}{2} \times 9.8\times 1.0642^2$

$h=5.3701\ m$

Now this height should be the maximum height of the tossed apple where its velocity becomes zero.

$v'^2=u'^2-2g.h$

$0^2=u'^2-2\times 9.8\times 5.3701$

$u'=10.259\ m.s^{-1}$ is the initial velocity of tossing the apple.

Time taken to reach this height:

$v'=u'-g.t'$

$0=10.259-9.8\times t'$

$t'=1.0469\ s$

We observe that $t>t'$ hence the time after the launch of the projectile after which the apple should be tossed is:

$\Delta t=t-t'$

$\Delta t=1.0642-1.0469$

$\Delta t=0.0173\ s$

8 0

3 years ago

"Determine the magnitude of the net force of gravity acting on the Moon during an eclipse when it is directly between Earth and

spayn [35]

Answer:

Net force = 2.3686 × 10^(20) N

Explanation:

To solve this, we have to find the force of the earth acting on the moon and the force of the sun acting on the moon and find the difference.

Now, from standards;

Mass of earth;M_e = 5.98 × 10^(24) kg

Mass of moon;M_m = 7.36 × 10^(22) kg

Mass of sun;M_s = 1.99 × 10^(30) kg

Distance between the sun and earth;d_se = 1.5 × 10^(11) m

Distance between moon and earth;d_em = 3.84 × 10^(8) m

Distance between sun and moon;d_sm = (1.5 × 10^(11)) - (3.84 × 10^(8)) = 1496.96 × 10^(8) m

Gravitational constant;G = 6.67 × 10^(-11) Nm²/kg²

Now formula for gravitational force between the earth and the moon is;

F_em = (G × M_e × M_m)/(d_em)²

Plugging in relevant values, we have;

F_em = (6.67 × 10^(-11) × 5.98 × 10^(24) × 7.36 × 10^(22))/(3.84 × 10^(8))²

F_em = 1.9909 × 10^(20) N

Similarly, formula for gravitational force between the sun and moon is;

F_sm = (G × M_s × M_m)/(d_sm)²

Plugging in relevant values, we have;

F_se = (6.67 × 10^(-11) × 1.99 × 10^(30) ×

7.36 × 10^(22))/(1496.96 × 10^(8))²

F_se = 4.3595 × 10^(20) N

Thus, net force = F_se - F_em

Net force = (4.3595 × 10^(20) N) - (1.9909 × 10^(20) N) = 2.3686 × 10^(20) N

8 0

2 years ago

How does your Liver work with your Bladder in your body?

andreyandreev [35.5K]

Answer:

The liver filters the blood of toxins before it continues through the body. It also removes excess sugar from the arteries. Your kidney then further filters the blood for water-soluble waste and toxins to be excreted and finally, your bladder gets read of that water-soluble waste.

6 0

2 years ago

A sound from a source has an intensity of 270 dB when it is 1 m from the source.

Rufina [12.5K]

Remember that sound intensity decreases in inverse proportion to the distance squared. So, to solve this we are going to use the inverse square formula: $\frac{I_{2} }{I_{1}}= (\frac{d{2} }{d_{1}})^2$
where
$I_{2}$ is the intensity at distance 2
$I_{1}$ is the intensity at distance 1
$d_{2}$ is distance 2
$d_{1}$ is distance 1

We can infer for our problem that $I_{1}=270$ , $d_{1}=1$ , and $d_{2}=3$ . Lets replace those values in our formula to find $I_{2}$ :
$\frac{I_{2} }{I_{1}}= (\frac{d{2} }{d_{1}})^2$
$\frac{I_{2} }{270} =( \frac{1}{3} )^2$
$\frac{I_{2} }{270} = \frac{1}{9}$
$I_{2}= \frac{270}{9}$
$I_{2}=30$ dB

We can conclude that the intensity of the sound when is 3 m from the source is 30 dB.

8 0

3 years ago

A skateboarder is moving at 2.00m/s and increases his velcity at a rate of 1.6m/s/s for 6.0 seconds. What is the displacement of

MrRissso [65]

Answer:

S = 40.8m

Explanation:

Given the following data;

Initial velocity, u = 2m/s

Acceleration, a = 1.6m/s²

Time, t = 6secs

Required to find the displacement

Displacement, S = ?

The displacement of an object is given by the second equation of motion;

$S = ut + \frac {1}{2}at^{2}$

Where;

S represents the displacement measured in meters.

u represents the initial velocity measured in meters per seconds.

t represents the time measured in seconds.

a represents acceleration measured in meters per seconds square.

Substituting into the equation, we have;

$S = 2*6 + \frac {1}{2}*(1.6)*6^{2}$

$S = 12 + 0.8*36$

$S = 12 + 28.8$

S = 40.8m

Therefore, the displacement of the skateboarder during this game is 40.8 meters.

7 0

3 years ago