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solniwko [45]
2 years ago
10

If you want the ball to land in your hand when it comes back down, should you toss the ball straight upward, in a forward direct

ion, or in a backward direction, relative to your body
Physics
1 answer:
denpristay [2]2 years ago
7 0

Straight upward

the ball moves in the forward direction with your walking speed at all times. If you want the ball to land in your hand when it comes back down, you should toss the ball straight upward.

<h3>What is Projectile motion ?</h3>

Projectile motion is the motion of an object thrown (projected) into the air.

  • After the initial force that launches the object, it only experiences the force of gravity. The object is called a projectile, and its path is called its trajectory

  • A projectile can be a thrown ball, a bullet or a springboard diver ... Except for air resistance, the forward velocity of any projectile is constant and is equal to the initial velocity when it was released.

Learn more about Projectile motion here:

brainly.com/question/27116954

#SPJ4

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A capacitor in a single-loop RC circuit is charged to 85% of its final potential difference in 2.4 s. What is the time constant
atroni [7]

Answer:

The  time constant is  \tau  = 1.265 s

Explanation:

From the question we are told that

     the time take to charge is  t = 2.4 \  s

The mathematically representation for voltage potential of a capacitor at different time is

        V  =  V_o  - e^{-\frac{t}{\tau} }

Where  \tau  is the time constant  

           V_o is the potential of the capacitor when it is full

     So  the capacitor potential will be  100%  when it is full thus  V_o  =100%  =  1  

and from the  question we are told that the  at the given time the potential of the capacitor is 85% = 0.85 of its final potential so

      V  = 0.85

Hence

     0.85 =  1 -  e^{-\frac{2.4}{\tau } }

       - {\frac{2.4}{\tau } }  =  ln0.15

        \tau  = 1.265 s

     

7 0
3 years ago
Select the correct answer.
VLD [36.1K]

Answer:O – H

Explanation:

8 0
3 years ago
A roller coaster pushes a 25 kg person upward with a force of 300 N. What is the acceleration?
vlada-n [284]
300N/25 kg= divide them for the answer
8 0
3 years ago
Zero, a hypothetical planet, has a mass of 5.3 x 1023 kg, a radius of 3.3 x 106 m, and no atmosphere. A 10 kg space probe is to
Andrej [43]

(a) 3.1\cdot 10^7 J

The total mechanical energy of the space probe must be constant, so we can write:

E_i = E_f\\K_i + U_i = K_f + U_f (1)

where

K_i is the kinetic energy at the surface, when the probe is launched

U_i is the gravitational potential energy at the surface

K_f is the final kinetic energy of the probe

U_i is the final gravitational potential energy

Here we have

K_i = 5.0 \cdot 10^7 J

at the surface, R=3.3\cdot 10^6 m (radius of the planet), M=5.3\cdot 10^{23}kg (mass of the planet) and m=10 kg (mass of the probe), so the initial gravitational potential energy is

U_i=-G\frac{mM}{R}=-(6.67\cdot 10^{-11})\frac{(10 kg)(5.3\cdot 10^{23}kg)}{3.3\cdot 10^6 m}=-1.07\cdot 10^8 J

At the final point, the distance of the probe from the centre of Zero is

r=4.0\cdot 10^6 m

so the final potential energy is

U_f=-G\frac{mM}{r}=-(6.67\cdot 10^{-11})\frac{(10 kg)(5.3\cdot 10^{23}kg)}{4.0\cdot 10^6 m}=-8.8\cdot 10^7 J

So now we can use eq.(1) to find the final kinetic energy:

K_f = K_i + U_i - U_f = 5.0\cdot 10^7 J+(-1.07\cdot 10^8 J)-(-8.8\cdot 10^7 J)=3.1\cdot 10^7 J

(b) 6.3\cdot 10^7 J

The probe reaches a maximum distance of

r=8.0\cdot 10^6 m

which means that at that point, the kinetic energy is zero: (the probe speed has become zero):

K_f = 0

At that point, the gravitational potential energy is

U_f=-G\frac{mM}{r}=-(6.67\cdot 10^{-11})\frac{(10 kg)(5.3\cdot 10^{23}kg)}{8.0\cdot 10^6 m}=-4.4\cdot 10^7 J

So now we can use eq.(1) to find the initial kinetic energy:

K_i = K_f + U_f - U_i = 0+(-4.4\cdot 10^7 J)-(-1.07\cdot 10^8 J)=6.3\cdot 10^7 J

3 0
3 years ago
A force of 100N is applied to an area of 100mm².what is the pressure exerted on the area in N/m².​
Dmitry_Shevchenko [17]

Answer:

P = 1000000[Pa] = 1000 [kPa]

Explanation:

To solve this problem we must use the definition of pressure, which is equal to the relationship of force over area.

P=F/A

where:

P = pressure [Pa] (units of pascals)

F = force = 100 [N]

A = area = 100 [mm²]

But first we must convert the units from square millimeters to square meters.

A=100[mm^{2}]*\frac{1^{2} m^{2} }{1000^{2}mm^{2}  } =0.0001[m^{2} ]

Now replacing:

P=100/0.0001\\P=1000000[Pa]

3 0
3 years ago
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