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katrin2010 [14]
3 years ago
12

35 pts please help quick

Physics
1 answer:
Studentka2010 [4]3 years ago
8 0

yes the car is chaning direction throught the track, so it is accelerting.


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!!! 100 POINTS !!! PLEASE HELP !!!
hram777 [196]

Q1. Option 2: basketball


Q2: Newton's first law is <span>the </span>law<span> of inertia. </span>An object at rest stays at rest and an object in motion stays in motion.

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<span>Q3. A basketball sitting on the floor stays there and a basketball rolling on court keeps on rolling.</span>

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<span>Q4 Second law says acceleration is dependent upon net force and mass of the object.</span>


Q5. Basketball accelerates when a player tries to dunk it with both hands.


<span>Q6. Third law says f<span>or every action, there is an equal and opposite reaction.</span></span>

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<span><span>Q7. As a player dribbles, the force the basketball hits the floor with is the same as the force from the floor on the ball. That is why the ball bounces back up in air.</span></span>

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8 0
3 years ago
Read 2 more answers
How much heat must be removed from 456 g of water at 25.0°C to change it into ice at - 10.0°C?
Svet_ta [14]

Answer:

229,098.96 J

Explanation:

mass of water (m) = 456 g = 0.456 kg

initial temperature (T) = 25 degrees

final temperature (t) = - 10 degrees

specific heat of ice = 2090 J/kg

latent heat of fusion =33.5 x 10^(4) J/kg

specific heat of water = 4186 J/kg

for the water to be converted to ice it must undergo three stages:

  • the water must cool from 25 degrees to 0 degrees, and the heat removed would be Q = m x specific heat of water x change in temp

        Q = 0.456 x 4186 x (25 - (-10)) = 66808.56 J

  • the water must freeze at 0 degrees, and the heat removed would be Q = m x specific heat of fusion x change in temp

         Q = 0.456 x 33.5 x 10^(4) = 152760 J

  • the water must cool further to -10 degrees from 0 degrees, and the heat removed would be Q = m x specific heat of ice x change in temp

        Q = 0.456 x 2090 x (0 - (-10)) = 9530.4 J

The quantity of heat removed from all three stages would be added to get the total heat removed.

Q total = 66,808.56 + 152,760 + 9,530.4 = 229,098.96 J

6 0
3 years ago
An electron is released from rest in a weak electric field given by vector E = -1.30 10-10 N/C ĵ. After the electron has travel
Murljashka [212]

Answer:

v =  6.45 10⁻³ m / s

Explanation:

Electric force is

               F = q E

Where q is the charge and E is the electric field

Let's use Newton's second law to find acceleration

                F- W = m a

               a = F / m - g

               a = q / m E  g

Let's calculate

               a = -1.6 10⁻¹⁹ / 9.1 10⁻³¹ (-1.30 10⁻¹⁰) - 9.8

              a = 0.228 10² -9.8

              a=  13.0 m / s²

Now we can use kinematics, knowing that the resting parts electrons

              v² = v₀² + 2 a y

              v =√ (0 + 2 13.0 1.6 10⁻⁶)

            v =  6.45 10⁻³ m / s

4 0
3 years ago
A ray of light incident in air strikes a rectangular glass block of refractive index 1.50, at an angle of incidence of 45°. Calc
balandron [24]

Answer:

Approximately 28^{\circ}.

Explanation:

The refractive index of the air n_{\text{air}} is approximately 1.00.

Let n_\text{glass} denote the refractive index of the glass block, and let \theta _{\text{glass}} denote the angle of refraction in the glass. Let \theta_\text{air} denote the angle at which the light enters the glass block from the air.

By Snell's Law:

n_{\text{glass}} \, \sin(\theta_{\text{glass}}) = n_{\text{air}} \, \sin(\theta_{\text{air}}).

Rearrange the Snell's Law equation to obtain:

\begin{aligned} \sin(\theta_{\text{glass}}) &= \frac{n_{\text{air}} \, \sin(\theta_{\text{air}})}{n_{\text{glass}}} \\ &= \frac{(1.00)\, (\sin(45^{\circ}))}{1.50} \\ &\approx 0.471\end{aligned}.

Hence:

\begin{aligned} \theta_{\text{glass}} &= \arcsin (0.471) \approx 28^{\circ}\end{aligned}.

In other words, the angle of refraction in the glass would be approximately 28^{\circ}.

7 0
2 years ago
A point charge of -3.0 x 10-5C is placed at the origin of coordinates. Find the electric field at the point 3. r= 50 m on the x-
Snezhnost [94]

Answer: -5×10-3

Explanation:

E=kq/r

4 0
2 years ago
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