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Zinaida [17]
2 years ago
5

How much thermal energy does it take to raise the temperature of 2.5 kg of

Physics
1 answer:
KonstantinChe [14]2 years ago
3 0

Answer:

390kJ

Explanation:

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A boy on a 1.9 kg skateboard initially at rest tosses a(n) 7.8 kg jug of water in the forward direction. if the jug has a speed
Tresset [83]
For this case we first think that the skateboard and the child are one body.
 We have then:
 1 = jug
 2 = skateboard + boy
 By conservation of the linear amount of movement:
 M1V1i + M2V2i = M1V1f + M2V2f
 Initial rest:
 v1i = v2i = 0
 0 = M1V1f + M2V2f
 Substituting values
 0 = (7.8) (3.2) + (M2) (- 0.65)
 0 = 24.96 + M2 (-0.65)
 -24.96 = (-0.65) M2
 M2 = (-24.96) / (- 0.65) = 38.4 kg
 Then, the child's mass is:
 M2 = Mskateboard + Mb
 Clearing:
 Mb = M2-Mskateboard
 Mb = 38.4 - 1.9
 Mb = 36.5 Kg
 answer:
 the boy's mass is 36.5 Kg
4 0
3 years ago
A national college researcher reported that 67%of students who graduated from high school in 2012 enrolled in college. Twenty-ni
dsp73

Answer:

Explanation:

Theorem of Binomial Distribution will apply here.

n = 29 , p = .67 , q = 0.33

mean = np = 29 x .67 = 19.43

Standard Deviation = √npq

= √29 x .67 x .33

= √6.4

= 2.53

=

4 0
3 years ago
A flywheel is a solid disk that rotates about an axis that is perpendicular to the disk at its center. Rotating flywheels provid
Sergeeva-Olga [200]

Answer:

8.37*10^5 rpm

Explanation:

Given that rotational kinetic energy = 4.66*10^9J

Mass of the fly wheel (m) = 19.7 kg

Radius of the fly wheel (r) = 0.351 m

Moment of inertia (I) = \frac{1}{2} mr ^2

Rotational K.E is illustrated as (K.E)_{rt} = \frac{1}{2} I \omega^2

\omega = \sqrt{\frac{2(K.E)_{rt}}{I} }

\omega = \sqrt{\frac{2(KE)_{rt}}{1/2 mr^2} }

\omega = \sqrt{\frac{4(K.E)_{rt}}{mr^2} }

\omega = \sqrt{\frac{4*4.66*10^9J}{19.7kg*(0.351)^2} }

\omega = 87636.04

\omega = 8.76*10^4 rad/s

Since 1 rpm = \frac{2 \pi}{60}  rad/s

\omega = 8.76*10^4(\frac{60}{2 \pi})

\omega = 836518.38

\omega = 8.37 *10^5 rpm

3 0
3 years ago
What is rotation in your own words. and you have to be detailed
attashe74 [19]

Answer:

The action of rotating around something like for ex.. Child A stands in the middle of a room while Child B goes around Child A, Child B is rotating around Child A. Another ex. is; The earth rotates around the sun every 365 days the earth is rotating or going around the sun in a circle. So rotating to me is the act of rotating around something in any way shape or form.

PS: A middle schooler answered this so if you don't wanna believe me or think I'm wrong because I am younger you do you.

7 0
3 years ago
A 0.140-kg baseball is dropped from rest from a height of 1.8 m above the ground. It rebounds to a height of 1.4 m. What change
aliina [53]

Answer:

\Delta p=-1.56\ kg-m/s

Explanation:

It is given that,

Mass of the baseball, m = 0.14 kg

It is dropped form a height of 1.8 m above the ground. Let u is the velocity when it hits the ground. Using the conservation of energy as :

u=\sqrt{2gh}

h = 1.8 m  

u=\sqrt{2\times 9.8\times 1.8}

u = 5.93 m/s

Let v is the speed of the ball when it rebounds. Again using the conservation of energy to find it :

v=\sqrt{2gh'}

h' = 1.4 m  

v=-\sqrt{2\times 9.8\times 1.4}

v = -5.23 m/s

The change in the momentum of the ball is given by :

\Delta p=m(v-u)

\Delta p=0.14(-5.23-5.93)

\Delta p=-1.56\ kg-m/s

So, the change in the ball's momentum occurs when the ball hits the ground is 1.56 kg-m/s. Hence, this is the required solution.

4 0
3 years ago
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