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2 years ago

Hello guys! Can u please help me with physics. I translated it in English. Can yall help me please how much u can!!

1 answer:

6 0

1. Since the body is thrown vertically upward, the only force acting on it as it rises and falls is gravity, which causes a constant downward acceleration with magnitude g = 9.8 m/s². Because this acceleration is constant, we can use the formula

v² - u² = 2a ∆x

where

u = initial speed

v = final speed

a = acceleration

∆x = displacement

At its maximum height, some distance y above the point where the body is launched, the body has zero velocity, so

0² - (20 m/s)² = 2 (-9.8 m/s²) y

Solve for y :

y = (20 m/s)² / (2 (9.8 m/s²)) ≈ 20.4 m

2. Relative to the ground, the body's maximum height is 60 m + 20.4 m ≈ 80.4 m.

3. At any time t ≥ 0, the body's vertical velocity is given by

v = 20 m/s - gt

At the highest point, we have

0 = 20 m/s - (9.8 m/s²) t

and solving for t gives

t = (20 m/s) / (9.8 m/s²) ≈ 2.04 s

4. The body's height y above the ground at any time t ≥ 0 is given by

y = 60 m + (20 m/s) t - 1/2 gt²

Solve for t when y = 0 :

0 = 60 m + (20 m/s) t - 1/2 (9.8 m/s²) t²

Using the quadratic formula,

t = (-b + √(b² - 4ac)) / (2a)

(and omitting the negative root, which gives a negative solution) where a = -1/2 (9.8 m/s²), b = 20 m/s, and c = 60 m. You should end up with

t ≈ 6.09 s

5. At the time found in (4), the body's velocity is

v = 20 m/s - g (6.09 s) ≈ -39.7 m/s

Speed is the magnitude of velocity, so the speed in question is 39.7 m/s.

You might be interested in

Which forces tend to slow down an object

laila [671]

The answer is "friction and air resistance" gravity does some of the work by keeping the object from floating away, but friction and air resistance does the biggest part. Friction is how rough the ground it meaning on tile, dirt, grass, etc... that would slow down the object and air resistance is the gravity pushing on the object also making it stop.

Hope this helps!

8 0

3 years ago

In a nuclear reactor 1 g of mass is converted into energy. How much energy in Joules is produced?

KengaRu [80]

Answer:

3 x 10^5 J

Explanation:

mass of substance, m = 1 g = 0.001 kg

Velocity of light, c = 3 x 10^8 m/s

According to the Einstein mass energy equivalence, the energy associated with the mass is given by

E = m c^2

E = 0.001 x 3 x 10^8

E = 3 x 10^5 J

5 0

3 years ago

A 0.00275 kg air‑inflated balloon is given an excess negative charge q1 =−3.50×10−8 C by rubbing it with a blanket. It is found

kati45 [8]

Answer:

1) $\rm q_2$ is<u> positive.</u>

<u></u>

2) $\rm q_2=4.56\times 10^{-10}\ C.$

Explanation:

<u></u>

The charged rod is held above the balloon and the weight of the balloon acts in downwards direction. To balance the weight of the balloon, the force on the balloon due to the rod must be directed along the upwards direction, which is only possible when the rod exerts an attractive force on the balloon and the electrostatic force on the balloon due to the rod is attractive when the polarities of the charge on the two are different.

Thus, In order for this to occur, the polarity of charge on the rod must be positive, i.e., $\rm q_2$ is <u>positive.</u>

<u></u>

<u></u>

<u>Given:</u>

Mass of the balloon, m = 0.00275 kg.
Charge on the balloon, $\rm q_1 = -3.50\times 10^{-8}\ C.$
Distance between the rod and the balloon, d = 0.0640 m.
Acceleration due to gravity, $\rm g = 9.81\ m/s^2.$

In order to balloon to be float in air, the weight of the balloom must be balanced with the electrostatic force on the balloon due to rod.

Weight of the balloon, $\rm W = mg = 0.00275\times 9.81=2.70\times 10^{-2}\ N.$

The magnitude of the electrostatic force on the balloon due to the rod is given by

$\rm F_e = \dfrac{1}{4\pi \epsilon_o}\dfrac{|q_1||q_2|}{d^2}.$

$\rm \dfrac{1}{4\pi \epsilon_o}$ is the Coulomb's constant.

For the elecric force and the weight to be balanced,

$\rm F_e = W\\\dfrac{1}{4\pi \epsilon_o}\dfrac{|q_1||q_2|}{d^2}=W\\8.99\times 10^9\times \dfrac{3.50\times10^{-8}\times |q_2| }{0.0640^2}=2.70\times 10^{-2}\\|q_2| = \dfrac{2.70\times 10^{-2}\times 0.00640^2}{8.99\times 10^9\times 2.70\times 10^{-7}}=4.56\times 10^{-10}\ C.$

3 0

2 years ago

(will give brainliest to the correct person!)

Neporo4naja [7]

Answer:

Explanation:

When a force hits something, an equal amount of force is exerted back on it.

3 0

3 years ago

Read 2 more answers

A long, straight wire lies in the plane of a circular coil with a radius of 0.018 m. the wire carries a current of 5.6 a and is

iris [78.8K]

(a) The net flux through the coil is zero.
In fact, the magnetic field generated by the wire forms concentric circles around the wire. The wire is placed along the diameter of the coil, so we can imagine as it divides the coil into two emisphere. Therefore, the magnetic field of the wire is perpendicular to the plane of the coil, but the direction of the field is opposite in the two emispheres. Since the two emispheres have same area, then the magnetic fluxes in the two emispheres are equal but opposite in sign, and so they cancel out when summing them together to find the net flux.

(b) If the wire passes through the center of the coil but it is perpendicular to the plane of the wire, the net flux through the coil is still zero.
In fact, the magnetic field generated by the wire forms concentric lines around the wire, so it is parallel to the plane of the coil. But the flux is equal to
$\Phi = BA \cos \theta$
where $\theta$ is the angle between the direction of the magnetic field and the perpendicular to the plane of the coil, so in this case $\theta=90^{\circ}$ and so the cosine is zero, therefore the net flux is zero.

5 0

3 years ago