Ask question

Ask question

All categories

2 years ago

6

Which of the following are examples of projectile motion?

1 answer:

liubo4ka [24]2 years ago

5 0

Answer: A. a basketball being shot toward the basket

Explanation: The definition of projectile motion is the motion of an object thrown or projected into the air, subject to only the acceleration of gravity. So, the basketball is the object being thrown and the person throwing the ball is aiming it to go into the basket making that the path of trajectory. Hope that makes sense and helps!

You might be interested in

Lava that cools so quickly that ions do not have time to arrange themselves into crystals will form igneous rocks with a _____ t

Alborosie

Lava cools so quickly that ions do not have time to arrange themselves into crystals will form igneous rocks with a glassy texture. Lava is the substance that flows from the volcano's.

8 0

3 years ago

A heating element has a resistance of 180 Ω and draws a current of 1.60 A. What is its power?

iragen [17]

its either 288Watts or 112.5 Watts

5 0

3 years ago

Read 2 more answers

Consider the rocket in the image. Which unbalanced force accounts for the direction of the net force of the rocket ?

Wittaler [7]

Answer:

unbalenced force

Explanation:

make Brainliest plz

3 0

3 years ago

Read 2 more answers

Angular velocity in the z direction of a flywheel is w(t)=A + Bt2 The numerical values of the constants are A=2.75 and B=1.50. W

Ivanshal [37]

Answer:

α(0) = 0 rad/s²

α(5) = 15 rad/s²

Explanation:

The angular velocity of the flywheel is given as follows:

w(t) = A + B t²

where, A and B are constants.

Now, for the angular acceleration, we must take derivative of angular velocity with respect to time:

Angular Acceleration = α (t) = dw/dt

α(t) = (d/dt)(A + B t²)

α(t) = 2 B t

where,

B = 1.5

<u>AT t = 0 s</u>

α(0) = 2(1.5)(0)

<u>α(0) = 0 rad/s²</u>

<u></u>

<u>AT t = 5 s</u>

α(5) = 2(1.5)(5)

<u>α(5) = 15 rad/s²</u>

6 0

3 years ago

SashulF [63]

Answer:

a. Final velocity, V = 2.179 m/s.

b. Final velocity, V = 7.071 m/s.

Explanation:

<u>Given the following data;</u>

Acceleration = 0.500m/s²

a. To find the velocity of the boat after it has traveled 4.75 m

Since it started from rest, initial velocity is equal to 0m/s.

Now, we would use the third equation of motion to find the final velocity.

$V^{2} = U^{2} + 2aS$

Where;

V represents the final velocity measured in meter per seconds.
U represents the initial velocity measured in meter per seconds.
a represents acceleration measured in meters per seconds square.
S represents the displacement measured in meters.

Substituting into the equation, we have;

$V^{2} = 0^{2} + 2*0.500*4.75$

$V^{2} = 4.75$

Taking the square root, we have;

$V^{2} = \sqrt {4.75}$

<em>Final velocity, V = 2.179 m/s.</em>

b. To find the velocity if the boat has traveled 50 m.

$V^{2} = 0^{2} + 2*0.500*50$

$V^{2} = 50$

Taking the square root, we have;

$V^{2} = \sqrt {50}$

<em>Final velocity, V = 7.071 m/s.</em>

8 0

3 years ago