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2 years ago

A 36,287 kg truck has a momentum of 907,175 kg • . What is the truck’s velocity?

2 answers:

7 0

By definition,
Momentum = Mass * Velocity

Let v = the velocity of the truck, m/s
The mass of the truck is 36,287 kg.
The momentum is 907,175 (kg-m)/s.

Therefore
907,175 (kg-m)/s = (36287 kg)*(v m/s)
v = 907175/36287 = 25 m/s

Answer: 25 m/s

SIZIF [17.4K]2 years ago

3 0

Answer:

the answer is 25 m/s

Explanation:

hope this helped out

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You throw a ball of mass 1 kg straight up. You observe that it takes 2.2 s to go up and down, returning to your hand. Assuming w

Elina [12.6K]

Answer:

10.791 m/s

5.93505 m

Explanation:

m = Mass of ball

$v_f$ = Final velocity

$v_i$ = Initial velocity

$t_f$ = Final time

$t_i$ = Initial time

g = Acceleration due to gravity = 9.81 m/s²

From the momentum principle we have

$\Delta P=F\Delta t$

Force

$F=mg$

So,

$m(v_f-v_i)=mg(t_f-t_i)\\\Rightarrow v_i=v_f-g(t_f-t_i)\\\Rightarrow v_i=0-(-9.81)(1.1-0)\\\Rightarrow v_i=10.791\ m/s$

The speed that the ball had just after it left the hand is 10.791 m/s

As the energy of the system is conserved

$K_i=U\\\Rightarrow \dfrac{1}{2}mv_i^2=mgh\\\Rightarrow h=\dfrac{v_i^2}{2g}\\\Rightarrow h=\dfrac{10.791^2}{2\times 9.81}\\\Rightarrow h=5.93505\ m$

The maximum height above your hand reached by the ball is 5.93505 m

5 0

2 years ago

Find earth's approximate mass from the fact that the moon orbits earth in an average time of 27.3 days at an average distance of

Aleks [24]

We can solve the problem by using Kepler's third law, which states:

$\frac{4 \pi^2}{T^2}=\frac{GM}{r^3}$

where T is the period of revolution of the Moon around the Earth, G is the gravitational constant, M the Earth's mass and r the average distance between Earth and Moon.

Using the data of the problem:

$T=27.3 d \cdot 24 \cdot 60 \cdot 60 = 2358720 s=2.36 \cdot 10^6 s$

$r=384000 km=3.84 \cdot 10^8 m$

We can re-arrange the equation and find the Earth's mass:

$M=\frac{4 \pi^2 r^3}{GT^2}=\frac{4 \pi^2 (3.84 \cdot 10^8 m)^3}{(6.67 \cdot 10^{-11})(2.36 \cdot 10^6 s)^2}=6.0 \cdot 10^{24} kg$

4 0

2 years ago

No official standard exists for extinguisher-cylinder colors. that said, a red cylinder is typically associated with which type

Lorico [155]

Fire extinguishers are typically red. Goes along with the red fire trucks and red being a color of emergency. While there isn't any standard, if we decided to change the colors of the fire extinguisher, people would be very confused and given the context, it could cost people's lives, so I'm thinking red will be the color of fire extinguishers for a while to come.
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8 0

2 years ago

Which combination of initial horizontal velocity, (vh) and initial vertical velocity, (vv) results in the greatest horizontal ra

Naya [18.7K]

If an object is projected with vertical speed given as

$v_v$

now the time of flight of the object that time in which it comes back on ground

$\Delta y = v_v t + \frac{1}{2}(-g)t^2$

now here we will have

$0 = v_v t - \frac{1}{2}gt^2$

$t = \frac{2v_v}{g}$

now the range of projectile is given as

$R = horizontal\: speed \times time$

$R = v_h(\frac{2v_v}{g}$

now here we know that

$v_v = v_0 sin\theta$

$v_h = v_0 cos\theta$

now the range is given as

$R = \frac{2(vsin\theta)(vcos\theta)}{g}$

$R = \frac{v^2sin2\theta}{g}$

now in order to have maximum range we can say

$sin2\theta = 1$

$2\theta = 90^0$

so we will have

$\theta = 45 ^0$

so now we can say

$v_v = v_h = \frac{v_0}{\sqrt2}$

so both speed must be same to have maximum horizontal range

8 0

2 years ago

Suppose you are asked to find the area of a rectangle that is 2.1-cm wide by 5.6-cm long. Your calculator answer would be 11.76

kicyunya [14]

If I’m understanding the question then 11.76cm^2 in two sig figs is 12cm^2. I don’t see a part c to this question though so if I’m not giving you the answer you need please advise and I will answer what you need! Thank you!!

8 0

2 years ago

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