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Alex
2 years ago
15

Momentum is a vector quantity because it includes velocity, which is also a

Physics
1 answer:
zvonat [6]2 years ago
7 0

Answer: A; True

Explanation: Momentum is known to be a vector quality, and thus has been proven by modern scientists and resulting in this answer being true.

Hope this helps <3

Stay safe, stay warm

-Carrie

Ps. it would mean a lot if you marked brainliest (=

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A 1 800-kg pile driver is used to drive a steel I-beam into the ground. The pile driver falls 4.60 m before coming into contact
N76 [4]

Answer:

F = 614913.88 N

Explanation:

We are given;

Mass of pile driver; m = 1800 kg

Height of fall of pole driver; h = 4.6 m

Depth driven into beam; d = 13.6 cm = 0.136 m

Now, from energy equations and applying to this question, we can write that;

Workdone = Change in potential energy

Formula for workdone is; W = F × d

While the average potential energy here is; W = mg(h + d)

Thus;

Fd = mg(h + d)

Where F is the average force exerted by the beam on the pile driver while in bringing it to rest.

Making F the subject, we have;

F = mg(h + d)/d

F = 1800 × 9.81 × (4.6 + 0.136)/0.136

F = 614913.88 N

7 0
3 years ago
DUE BY MIDNIGHT
olga2289 [7]

Answer:

Option D. 1000 J.

Explanation:

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

Force (F) applied = 200 N

Distance (s) = 5 m

Time (t) = 10 s

Workdone (Wd) =?

Workdone (Wd) is simply defined as the product of force (F) and distance (s) moved in the direction of the force. Mathematically, it is expressed as:

Wd = F × s

With the above formula, we can calculate the Workdone as illustrated below:

Force (F) applied = 200 N

Distance (s) = 5 m

Workdone (Wd) =?

Wd = F × s

Wd = 200 × 5

Wd = 1000 J

Thus, the Workdone is 1000 J

8 0
2 years ago
Select the correct answer.
LekaFEV [45]
D is not a scientific experiment.
4 0
3 years ago
A projectile is launched with speed v0 at an angle of θ0 above the horizontal. Find an expression for the maximum height it reac
Step2247 [10]

Answer:

h= \frac{(v_{o})^{2} sin^{2} \theta o }{2g}

Explanation:

The parabolic movement results from the composition of a uniform rectilinear motion (horizontal ) and a uniformly accelerated rectilinear motion of upward or downward motion (vertical ).

The equation of uniform rectilinear motion (horizontal ) for the x axis is :

x = xi + vx*t   Equation (1)

Where:  

x: horizontal position in meters (m)

xi: initial horizontal position in meters (m)

t : time (s)

vx: horizontal velocity  in m/s  

The equations of uniformly accelerated rectilinear motion of upward (vertical ) for the y axis  are:

y= y₀+(v₀y)*t - (1/2)*g*t² Equation (2)

vfy= v₀y -gt Equation (3)

vfy²= v₀y²-2gH Equation (4)

Where:  

y: vertical position in meters (m)  

y₀ : initial vertical position in meters (m)  

t : time in seconds (s)

v₀y: initial  vertical velocity  in m/s  

vfy: final  vertical velocity  in m/s  

g: acceleration due to gravity in m/s²

H= hight that reaches the projectile above its starting point (m)

Data

v₀  : total initial speed

θ₀ : angle of v₀ above the horizontal.

g acceleration due to gravity

Calculation of the componentes x-y of the v₀

v₀x = v₀*cos θ₀

v₀y = v₀*sin θ₀

Calculation of the maximum hight that reaches the projectile above its starting point

When the projectile reaches its maximum height (h), vy = 0:

in the Equation (4)

vfy²= v₀y²-2gh

0= v₀y²-2gh

2gh= v₀y²h= \frac{(v_{o})^{2}sin^{2} \theta o }{2g}

2gh=  (v₀*sin θ₀)²

h= \frac{v_{o}^{2} sin^{2} \theta o }{2g}

7 0
2 years ago
Joey, whose mass is m = 36 kg, stands at rest at the outer edge of a frictionless merry-go-round with the mass M = 300 kg and th
Evgen [1.6K]

Answer:

\omega=0.24\ rad.s^{-1}

Explanation:

Given:

mass of person, m=36\ kg

mass of merry go-round, M=300\ kg

radius of merry go-round, R=2\ m

velocity of the person running, v=4\ m.s^{-1}

<u>We consider merry go-round as a ring:</u>

Now the moment of inertial of the ring is given as,

I=M.R^2

I=300\times 2^2

I=1200\ kg.m^{-2}

<u>Moment of inertia of the person considering as a point mass:</u>

I_p=m.R^2

I_p=36\times 2^2

I_p=144\ kg.m^2

<u>Now according to the conservation of angular momentum:</u>

I.\omega=I_p.\omega_p

where:

\omega = angular velocity of the merry-go-round

\omega_p= angular velocity of the person running

1200\times \omega=144\times \frac{v}{R}

\omega=\frac{144}{1200} \times \frac{4}{2}

\omega=0.24\ rad.s^{-1}

4 0
3 years ago
Read 2 more answers
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