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3 years ago

How is velocity different from speed?

Physics

2 answers:

Dimas [21]3 years ago

7 0

Answer:

B. Velocity includes direction.

Explanation:

edgenuity2020

enyata [817]3 years ago

5 0

Velocity, unlike speed, includes a direction.

Velocity is a vector quantity which is defined by magnitude and direction.

Speed is a scalar quantity. It is the rate at which an object moves regardless of which direction.

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A ball is thrown with a speed of 20 m/s at an angle of 40o above the horizontal from the top of a 22-m tall building.

MrRissso [65]

Answer:

(a). The distance is 49.79 m.

(b). The speed of the ball is 24.39 m/s.

Explanation:

Given that,

Speed = 20 m/s

Angle = 40°

Height = 22 m

Time = 3.25 sec

(a). We need to calculate the distance

Using formula of distance

$d=u\cos\theta\times t$

Put the value into the formula

$d=20\cos40\times3.25$

$d=49.79\ m$

(b). We need to calculate the horizontal velocity

Using formula of velocity

$v_{x}=u\cos\theta$

Put the value into the formula

$v_{x}=20\times\cos40$

$v_{x}=15.3\ m/s$

We need to calculate the vertical velocity

Using equation of motion

$v_{y}=u\sin\theta-gt$

Put the value into the formula

$v_{y}=20\sin40-9.8\times3.25$

$v_{y}=-19\ m/s$

Negative sign shows the opposite direction.

We need to calculate the speed of ball

Using formula of speed

$v=\sqrt{v_{x}^2+v_{y}^2}$

$v=\sqrt{(15.3)^2+(19)^2}$

$v=24.39\ m/s$

Hence, (a). The distance is 49.79 m.

(b). The speed of the ball is 24.39 m/s.

4 0

3 years ago

A 5.00 g object moving to the right at 20.0 cm/s makes an elastic head-on collision with a 10.0 g object that is initially at re

marusya05 [52]

Answer: a) 6.67cm/s b) 1/2

Explanation:

According to law of conservation of momentum, the momentum of the bodies before collision is equal to the momentum of the bodies after collision. Since the second body was initially at rest this means the initial velocity of the body is "zero".

Let m1 and m2 be the masses of the bodies

u1 and u2 be their velocities respectively

m1 = 5.0g m2 = 10.0g u1 = 20.0cm/s u2 = 0cm/s

Since momentum = mass × velocity

The conservation of momentum of the body will be

m1u1 + m2u2 = (m1+m2)v

Note that the body will move with a common velocity (v) after collision which will serve as the velocity of each object after collision.

5(20) + 10(0) = (5+10)v

100 + 0 = 15v

v = 100/15

v = 6.67cm/s

Therefore the velocity of each object after the collision is 6.67cm/s

b) kinectic energy of the 10.0g object will be 1/2MV²

= 1/2×10×6.67²

= 222.44Joules

kinectic energy of the 5.0g object will be 1/2MV²

= 1/2×5×6.67²

= 222.44Joules

= 111.22Joules

Fraction of the initial kinetic transferred to the 10g object will be

111.22/222.44

= 1/2

3 0

3 years ago

Why doesn't every planet have a moon?

GalinKa [24]

Answer:

Up first are Mercury and Venus. Neither of them has a moon. Because Mercury is so close to the Sun and its gravity, it wouldn't be able to hold on to its own moon. Any moon would most likely crash into Mercury or maybe go into orbit around the Sun and eventually get pulled into it.

5 0

2 years ago

What is playing catch? ( in terms of the force of the air interacting with ball )

Ugo [173]

Playing catch involves the ball moving in one direction in the air when you exert force. When the ball is in air, there is a gravitational force and a little it of backward air resistance until the other person catches it.

6 0

3 years ago

A football kicked in front of a goal post at an angle of 45 degree to the ground just clear the top by of the post 3m high. Calc

Firlakuza [10]

Answer:

A. 10.84 m/s

B. 1.56 s

Explanation:

From the question given above, the following data were obtained:

Angle of projection (θ) = 45°

Maximum height (H) = 3 m

Acceleration due to gravity (g) = 9.8 m/s²

Velocity of projection (u) =?

Time (T) taken to hit the ground again =?

A. Determination of the velocity of projection.

Angle of projection (θ) = 45°

Maximum height (H) = 3 m

Acceleration due to gravity (g) = 9.8 m/s²

Velocity of projection (u) =?

H = u²Sine²θ / 2g

3 = u²(Sine 45)² / 2 × 9.8

3 = u²(0.7071)² / 19.6

Cross multiply

3 × 19.6 = u²(0.7071)²

58.8 = u²(0.7071)²

Divide both side by (0.7071)²

u² = 58.8 / (0.7071)²

u² = 117.60

Take the square root of both side

u = √117.60

u = 10.84 m/s

Therefore, the velocity of projection is 10.84 m/s.

B. Determination of the time taken to hit the ground again.

Angle of projection (θ) = 45°

Velocity of projection (u) = 10.84 m/s

Time (T) taken to hit the ground again =?

T = 2uSine θ /g

T = 2 × 10.84 × Sine 45 / 9.8

T = 21.68 × 0.7071 / 9.8

T = 1.56 s

Therefore, the time taken to hit the ground again is 1.56 s.

7 0

3 years ago