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Ray Of Light [21]
3 years ago
10

A ball is launched from the ground with a horizontal speed of 30 m/s and a vertical speed of 30 m/s. what is the ball's horizont

al acceleration at the top of its trajectory?
Physics
1 answer:
andreyandreev [35.5K]3 years ago
6 0

the horizontal acceleration of the ball at the top of the trajectory is zero.

in case of projectile motion,only the force of gravity acts on the object along the vertical. There are no horizontal forces acting on the object.

now force = m a

m= mass

a= horizontal acceleration

f= horizontal force=0

so a= F/m

a=0/m=0

so the horizontal acceleration at all the points including the top of the trajectory is zero.

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Current that moves in one direction from negative to positive. May be created by a battery. Is generally NOT found in U.S. Elect
Karolina [17]

Answer:

C. D.C.

Explanation:

The current that is being described here is D.C. or direct current. It is the D.C. that moves in one direction from negative to positive. May be created by a battery. It is different from the A.C.( alternating current whose polarity changes regularly). It is A.C. that is used in electrical outlets and not D.C.So, the current option is C.

3 0
3 years ago
Read 2 more answers
If the potential across two parallel plates, separated by 4.0 cm, is 15.0 V, what is the electric field strength in volts per me
abruzzese [7]

Field strength = (15 V) / (4 cm)

Field strength = (15 V) / (0.04 meter)

Field strength = (15/0.04) (volts/meter)

<em>Field strength = 375 volts/meter </em>

3 0
3 years ago
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If an automobile engine delivers 42.0 hp of power, how much time will it take for the engine to do 6.20 â 105 j of work? (hint:
Elena L [17]
To be able to answer this item, we are to calculate the power that the machine could deliver from hp to kW. 

      (45 hp)(746 W/1 hp) = 33570 W

Power is the amount of energy delivered at a certain period. 

             t = (6.20 x 10^2 J)/ (33570 kJ/s)

             t = 0.01845 s
7 0
3 years ago
Twin A makes a one way trip at 0.6c to a star 12 light years away while twin B stays on Earth. Each twin sends the other a signa
hram777 [196]

Answer:

8 signals received by twin A during the trip.

Explanation:

Given that,

Distance = 12 light year

Speed = 0.6 c

Time = 1 year

We need to calculate the time by A

Using formula of time

T=t\sqrt{\dfrac{1+\dfrac{v}{c}}{1-\dfrac{v}{c}}}

Put the value into the formula

T=1\sqrt{\dfrac{1+0.6}{1-0.6}}

T=2\ years

Similarly,

The expression for distance cover by A

D=d\sqrt{1-\dfrac{v^2}{c^2}}

D=12\sqrt{1-(0.6)^2}

D=9.6\ ly

We need to calculate the time

Using formula of time

t=\dfrac{D}{v}

t=\dfrac{9.6}{0.6}

t=16\ years

We need to calculate the signals received by twin A

Using formula for number of signals

n=\dfrac{t}{T}

Put the value into the formula

n=\dfrac{16}{2}

n=8\ signals

Hence, 8 signals received by twin A during the trip.

7 0
3 years ago
A rocket is fired straight up. It contains two stages (Stage 1 and Stage 2) of solid rocket fuel that are designed to burn for 1
Archy [21]

Answer:

a)  y= 3.5 10³ m, b)   t = 64 s

Explanation:

a) For this exercise we use the vertical launch kinematics equation

Stage 1

          y₁ = y₀ + v₀ t + ½ a t²

          y₁ = 0 + 0 + ½ a₁ t²

Let's calculate

         y₁ = ½ 16 10²

         y₁ = 800 m

At the end of this stage it has a speed

        v₁ = vo + a₁ t₁

        v₁ = 0 + 16 10

        v₁ = 160 m / s

Stage 2

        y₂ = y₁ + v₁ (t-t₀) + ½ a₂ (t-t₀)²

        y₂ = 800 + 150 5 + ½ 11 5²

        y₂ = 1092.5 m

Speed ​​is

        v₂ = v₁ + a₂ t

        v₂ = 160 + 11 5

        v₂ = 215 m / s

The rocket continues to follow until the speed reaches zero (v₃ = 0)

         v₃² = v₂² - 2 g y₃

         0 = v₂² - 2g y₃

         y₃ = v₂² / 2g

         y₃ = 215²/2 9.8

         y₃ = 2358.4 m

The total height is

          y = y₃ + y₂

          y = 2358.4 + 1092.5

          y = 3450.9 m

           y= 3.5 10³ m

b) Flight time is the time to go up plus the time to go down

Let's look for the time of stage 3

          v₃ = v₂ - g t₃

          v₃ = 0

          t₃ = v₂ / g

          t₃ = 215 / 9.8

          t₃ = 21.94 s

The time to climb is

          t_{s} = t₁ + t₂ + t₃

          t_{s} = 10+ 5+ 21.94

          t_{s} = 36.94 s

The time to descend from the maximum height is

          y = v₀ t - ½ g t²

When it starts to slow down it's zero

         y = - ½ g t_{b}²

         t_{b}  = √-2y / g

         

        t_{b} = √(- 2 (-3450.9) /9.8)

        t_{b} = 26.54 s

Flight time is the rise time plus the descent date

        t = t_{s} + t_{b}

        t = 36.94 + 26.54

        t =63.84 s

        t = 64 s

3 0
3 years ago
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