I need help the option for D is the golfball isnt accelerating

Two boxes of masses 3M and 5M are attached by a massless rope. They are being pulled to the right with a constant force of P = 8

Makovka662 [10]

Answer:

a) about 14.577 kg

b) 300 N

Explanation:

b) In order for the acceleration to be the same for each mass, the 800 N force must be divided between the boxes in proportion to their mass. That is, the net force on the 5M mass must be 5/8 of the total force, or 500 N. Then the tension in the rope is 800 N -500 N = 300 N, which is 3/8 of 800 N.

Tension: 300 N

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a) The total mass is 8M, and the total normal force on the floor is ...

F = ma = (8M)(9.8 m/s^2)

The friction force is 0.7 times this, and is equal to the 800 N force pulling on the boxes.

800 N = (8M)(9.8 m/s^2)(0.7)

M = 800/(8·9.8·0.7) kg ≈ 14.577 kg

3 0

3 years ago

The height (in meters) of a projectile shot vertically upward from a point 3 m above ground level with an initial velocity of 21

frutty [35]

Answer:

a) The velocity of the projectile at 2 seconds after launch is 1.9 meters per second. The velocity of the projectile at 4 seconds after launch is -17.7 meters per second.

b) The projectile reaches maximum height 2.192 seconds after launch.

c) The maximum height of the projectile is 26.584 meters above ground.

d) The projectile will hit the ground at 4.523 seconds after launch.

e) The velocity of the projectile right before hitting the ground in -22.871 meters per second.

Explanation:

Complete statement of problem is: The height (in meters) of a projectile shot vertically upward from a point 3 m above ground level with an initial velocity of 21.5 meters per second is $h(t) = 3+21.5\cdot t-4.9\cdot t^{2}$ after t seconds. (Round your answers to two decimal places.) (a) Find the velocity after 2 seconds and after 4 seconds, (b) When does the projectile reach its maximum height? (c) What is the maximum height? (d) When does it hit the ground? (e) With what velocity does it hits the ground?

a) From Physics and Differential Calculus we remember that velocity is the first derivative of height. Hence, we need to differentiate the height function in time:

$v(t) = 21.5-9.8\cdot t$ (Eq. 1)

Where $v(t)$ is the velocity function, measured in meters per second.

Now we evaluate this function at given times:

t = 2 s.

$v(2) = 21.5-9.8\cdot (2)$

$v(2) = 1.9\,\frac{m}{s}$

The velocity of the projectile at 2 seconds after launch is 1.9 meters per second.

t = 4 s.

$v(4) = 21.5-9.8\cdot (4)$

$v(4) = -17.7\,\frac{m}{s}$

The velocity of the projectile at 4 seconds after launch is -17.7 meters per second.

b) Maximum height is reached when velocity of projectile is zero. We equalize velocity to zero and solve the expression for $t$ :

$21.5-9.81\cdot t = 0$

$t = 2.192\,s$

The projectile reaches maximum height 2.192 seconds after launch.

c) Maximum height is calculated by evaluating height function at the time found in b). That is:

$h(2.192) = 3+21.5\cdot (2.192)-4.9\cdot (2.192)^{2}$

$h (2.192) = 26.584\,m$

The maximum height of the projectile is 26.584 meters above ground.

d) In this case, we need to equalize the height function to zero and solve for $t$ . That is:

$3+21.5\cdot t-4.9\cdot t^{2} = 0$

Roots are found by means of Quadratic Formula:

$t_{1}\approx 4.523\,s$ and $t_{2}\approx -0.135\,s$

Only the first root offers a physically reasonable solution. Therefore, the projectile will hit the ground at 4.523 seconds after launch.

e) This can be found by evaluating velocity function at the time found in d):

$v(4.523) = 21.5-9.81\cdot (4.523)$

$v(4.523) = -22.871\,\frac{m}{s}$

The velocity of the projectile right before hitting the ground in -22.871 meters per second.

5 0

4 years ago

You are pulling a child in a wagon. The rope handle is inclined upward at a 60∘ angle. The tension in the handle is 20 N.

dem82 [27]

Angle (θ) = 60°
Force (F) = 20 N
Distance (s) = 200 m
Therefore, work done
= Fs Cos θ
= (20 × 200 × Cos 60°) J
= (20 × 200 × 1/2) J
= (20 × 100) J
= 2000 J

Answer:

2000 J

Hope you could get an idea from here.

Doubt clarification - use comment section.

6 0

2 years ago

Read 2 more answers

An 80kg person falls 60 m off of a waterfall. What is her GPE?

ElenaW [278]

Answer:

The gravitational potential energy of the man

= mass of the man(m) × gravitational acceleration(g) × height (h)

80 Kg × 9.8 m/s^2 × 60 m

80 × 9.8 x 60 ( kg ×m^2/s^2)

47040 Joules (ans)

Hope it helps

4 0

3 years ago

When an object accelerates, what about its motion changes? Question 1 options: Speed must change, but not velocity. Both speed a

QveST [7]

Answer:

The velocity must change but not speed.

Explanation:

Velocity is defined as the displacement by time. Whereas speed is expressed as the distance between two successive positions of the body to the time interval it took to travel.

Velocity, V = D / t m/s

Speed, s = d /t m/s

Velocity is a vector quantity that has a magnitude and direction.
The speed is a scalar quantity having only the magnitude.
At any instant of time, the magnitude of the velocity is always equal to the magnitude of the speed. The magnitude of velocity, |v | = magnitude of speed, |v |. The magnitude is always positive
The acceleration of a body is defined as the rate of change of velocity to time.

a = (v - u) / t m/s²

If a body is accelerating, It varies its velocity with respect to time.
In case of uniform circular motion, the speed remains constant, but the velocity changes continuously.

So, in the case of circular motion if an object accelerates, velocity must change but speed remains constant.

5 0

4 years ago