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2 years ago

A boy is standing motionless on a skateboard. He throws a basketball forward. Describe his motion.

Physics

1 answer:

weqwewe [10]2 years ago

8 0

Answer:

Random motion

Explanation:

If the boy throws the basketball forward while at a position on the skateboard, the motion of the ball will be a random motion since we are not told if the ball is moving on a straight line when thrown forward.

In this case, the boy will tend to move in the direction of the ball. Since the ball is moving in a random manner, the motion of the boy will also be a random motion.

A random motion is a motion of a body in a zig zag manner. It is also known as Brownian motion e.g motion of a buzzing mosquito, motion of a smoke coming out of a chimney etc.

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In the Daytona 500 auto race, a Ford Thunderbird and a Mercedes Benz are moving side by side down a straightaway at 71.0 m/s. Th

qaws [65]

Answer:

The distance between both cars is 990 m

Explanation:

The equations for the position and the velocity of an object moving in a straight line are as follows:

x = x0 + v0 * t + 1/2 * a * t²

v = v0 + a * t

where:

x = position of the car at time "t"

x0 = initial position

v0 = initial speed

t = time

a = acceleration

v = velocity

First let´s find how much time it takes the driver to come to stop (v = 0). We will consider the origin of the reference system as the point at which the driver realizes she must stop. Then x0 = 0

With the equation of velocity, we can obtain the acceleration and replace it in the equation of position, knowing that the position will be 250 m at that time.

v = v0 + a*t

v-v0 / t = a

0 m/s - 71.0 m/s / t =a

-71.0 m/s / t = a

Replacing in the equation for position:

x = v0* t +1/2 * a * t²

250 m = 71.0 m/s * t + 1/2 *(-71.0 m/s / t) * t²

250 m = 71.0 m/s * t - 1/2 * 71.0 m/s * t

250m = 1/2 * 71.0m/s *t

t = 2 * 250 m / 71.0 m/s = 7.04 s

It takes the driver 7.04 s to stop.

Then, we can calculate how much time it took the driver to reach her previous speed. The procedure is the same as before:

v = v0 + a*t

v-v0 / t = a now v0 = 0 and v = 71.0 m/s

(71.0 m/s - 0 m/s) / t = a

71.0 m/s / t =a

Replacing in the position equation:

x = v0* t +1/2 * a * t²

390 m = 0 m/s * t + 1/2 * 71.0 m/s / t * t² (In this case, the initial position is in the pit, then x0 = 0 because it took 390 m from the pit to reach the initial speed).

390m * 2 / 71.0 m/s = t

t = 11.0 s

In total, it took the driver 11.0s + 5.00 s + 7.04 s = 23.0 s to stop and to reach the initial speed again.

In that time, the Mercedes traveled the following distance:

x = v * t = 71.0 m/s * 23.0 s = 1.63 x 10³ m

The Thunderbird traveled in that time 390 m + 250 m = 640 m.

The distance between the two will be then:

distance between both cars = 1.63 x 10³ m - 640 m = 990 m.

3 0

2 years ago

Two objects are placed in thermal contact and are allowed to come to equilibrium in isolation. the heat capacity of object a is

Harman [31]

Given:
Ca = 3Cb (1)
where
Ca = heat capacity of object A
Cb = heat capacity f object B

Also,
Ta = 2Tb (2)
where
Ta = initial temperature of object A
Tb = initial temperature of object B.

Let
Tf = final equilibrium temperature of both objects,
Ma = mass of object A,
Mb = mass of object B.

Assuming that all heat exchange occurs exclusively between the two objects, then energy balance requires that
Ma*Ca*(Ta - Tf) = Mb*Cb*(Tf - Tb) (3)

Substitute (1) and (2) into (3).
Ma*(3Cb)*(2Tb - Tf) = Mb*Cb*(Tf - Tb)
3(Ma/Mb)*(2Tb - Tf) = Tf - Tb

Define k = Ma/Mb, the ratio f the masses.
Then
3k(2Tb - Tf) = Tf - Tb
Tf(1+3k) = Tb(1+6k)
Tf = [(1+6k)/(1+3k)]*Tb

Answer:
$T_{f} =( \frac{1+6k}{1+3k} )T_{b}= \frac{1}{2}( \frac{1+6k}{1+3k})T_{a}$
where
$k= \frac{M_{a}}{M_{b}}$

7 0

2 years ago

A vessel, divided into two parts by a partition, contains 4 mol of nitrogen gas at 75°C and 30 bar on one side and 2.5 mol of ar

Elena L [17]

Answer:

assume nitrogen is an ideal gas with cv=5R/2

assume argon is an ideal gas with cv=3R/2

n1=4moles

n2=2.5 moles

t1=75°C in kelvin t1=75+273

t1=348K

T2=130°C in kelvin t2=130+273

t2=403K

u=пCVΔT

U(N₂)+U(Argon)=0

putting values:

=>4x(5R/2)x(Tfinal-348)=2.5x(3R/2)x(T final-403)

by simplifying:

Tfinal=363K

6 0

2 years ago

A 6.0-μF capacitor charged to 50 V and a 4.0-μF capacitor charged to 34 V are connected to each other, with the two positive pla

ch4aika [34]

Answer:

5702.88 J or 5.7mJ

Explanation:

Given that :

C 1 = 6.0-μF

C 2 = 4.0-μF

V 1 = 50V

V 2 = 34V

Note that : Q = CV

Q 1 = C1 * V1

Q 1 = 50×6 = 300μC

Q 2 = 34×4 = 136μC

Parallel connection = C 1 + C 2

= 6+4 = 10μC

V = Qt/C

Where Qt = Q1+Q2

V = Q1+Q2/C

V = 300+136/10

V = 437/10

V = 43.6volts

Uc1 = 1/2×C1V^2

= 1/2 × 6μF × 43.6^2

= 1/2 × 6μF × 1900.96

= 3μF × 1900.96volts

= 5702.88J

= 5702.88J/1000

= 5.7mJ

4 0

2 years ago

8) A soccer goal is 2.44 m high. A player kicks the ball at a distance 10.0 m from the goal at an angle of 25.0°. The ball hits

White raven [17]

Answer:

The initial speed of the soccer ball is 16.38 m/s

Explanation:

given;

vertical distance y = 2.44 m

horizontal distance x = 10.0 m

angle of projection θ = 25.0°

Initial velocity has two components, Vₓ and V $_y$

Vₓ = V $_i$ cosθ

V $_y$ = V $_i$ sinθ

The horizontal distance = x = Vₓt + ¹/₂ ˣ g ˣ t², but g =0

x = Vₓt = V $_i$ cosθ *t

10 = V $_i$ cos25 *t

10 = 0.906V $_i$ *t

V $_i$ *t = 10/0.906 = 11.038 m

The vertical distance (g = - g, because it upward motion against gravity)

y = V $_y$ *t -¹/₂ ˣ g ˣ t²

2.44 = (V $_i$ sinθ)t - ¹/₂ ˣ 9.8 ˣ t²

2.44 = (V $_i$ *t)sinθ - ¹/₂ ˣ 9.8 ˣ t²

2.44 = (11.038)sin25° - 4.9t²

2.44 = (11.038)*0.4226 - 4.9t²

2.44 = 4.6647 - 4.9t²

4.9t² = 4.6647 - 2.44 = 2.2247

t² = 2.2247/4.9

t² = 0.454

t = √0.454

t = 0.674 s

Recall that V $_i$ *t = 11.038 m

V $_i$ *0.674 = 11.038 m, solve for V $_i$

V $_i$ = 11.038/0.674

V $_i$ = 16.38 m/s

Therefore, the initial speed of the soccer ball is 16.38 m/s

6 0

2 years ago