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NeX [460]
3 years ago
10

Speed is a measure of ___ over ____

Physics
2 answers:
Pavlova-9 [17]3 years ago
5 0

Ok so the answer is some what like Distance Divided BY Time? I think

Anarel [89]3 years ago
4 0

distance divided by time

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A student standing on a stationary skateboard tosses a textbook with a mass of mb = 1.35 kg to a friend standing in front of him
nataly862011 [7]

Answer:

a) u_c=0\ m.s^{-1}       &        m_c.v_c=m_b.v_b\times \cos\theta

b) v_c=0.0566\ m.s^{-1}

c) p_e=2.9218\ kg.m.s^{-1}

Explanation:

Given:

mass of the book, m_b=1.35\ kg

combined mass of the student and the skateboard, m_c=97\ kg

initial velocity of the book, v_b=4.61\ m.s^{-1}

angle of projection of the book from the horizontal, \theta=28^{\circ}

a)

velocity of the student before throwing the book:

Since the student is initially at rest and no net force acts on the student so it remains in rest according to the Newton's first law of motion.

u_c=0\ m.s^{-1}

where:

u_c= initial velocity of the student

velocity of the student after throwing the book:

Since the student applies a force on the book while throwing it and the student standing on the skate will an elastic collision like situation on throwing the book.

m_c.v_c=m_b.v_b\times \cos\theta

where:

v_c= final velcotiy of the student after throwing the book

b)

m_c.v_c=m_b.v_b\times \cos\theta

97\times v_c=1.35\times 4.61\cos28

v_c=0.0566\ m.s^{-1}

c)

Since there is no movement of the student in the vertical direction, so the total momentum transfer to the earth will be equal to the momentum of the book in vertical direction.

p_e=m_b.v_b\sin\theta

p_e=1.35\times 4.61\times \sin28^{\circ}

p_e=2.9218\ kg.m.s^{-1}

6 0
3 years ago
Un móvil se encuentra en la posición inicial x0 = 22 m, y se mueve con velocidad 5 m/s. Calcula la posición en la que se encuent
Brilliant_brown [7]

Answer:

La posición en la que se encuentra el móvil en el instante t = 30 s es 172 m.

Explanation:

El movimiento rectilíneo uniforme (MRU) es el movimiento que describe un cuerpo o partícula a través de una línea recta a velocidad constante.

La distancia recorrida,  x , por un móvil que tiene un MRU con un velocidad  v  durante el intervalo de tiempo  t  es:

x= x0 + v*t

donde x0 es la posición inicial.

En este caso:

  • x0= 22 m
  • v= 5 m/s
  • t= 30 s

Reemplazando:

x= 22 m + 5 m/s* 30 s

Resolviendo:

x= 22 m + 150 m

x= 172 m

<u><em>La posición en la que se encuentra el móvil en el instante t = 30 s es 172 m.</em></u>

8 0
3 years ago
An object travels at 15 meters in 10 seconds east. What is it’s velocity?
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A new jxo rib is excellent is.
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Nuclearissionoccursthroughmanydifferent pathways. For the ission of U-235 induced by a neutron, write a nuclear equation to form
ruslelena [56]

Answer: a) ^{235}_{92}\textrm{U}+^{1}_{0}\textrm{n} \rightarrow ^{87}_{35}\textrm {Br}+ ^{146}_{57}\La + 3^{1}_{0}n

b) ^{235}_{92}\textrm{U}+^1_0\textrm{n}\rightarrow ^{140}_{56}\textrm{Ba}+^{94}_{36}\textrm{Kr}+2^1_0\textrm{n}

Explanation:

A nuclear fission reaction is defined as the reaction in which a heavy nucleus splits into small nuclei along with release of energy.

a) The given reaction is ^{235}_{92}\textrm{U}+^{1}_{0}\textrm{n} \rightarrow ^{87}_{35}\textrm {Br}+ ^{146}_{57}\La + x^{1}_{0}n

Now,  as the mass on both reactant and product side must be equal:

235+1=87+146+x

x=3

Thus three neutrons are produced and nuclear equation will be: ^{235}_{92}\textrm{U}+^{1}_{0}\textrm{n} \rightarrow ^{87}_{35}\textrm {Br}+ ^{146}_{57}\La + 3^{1}_{0}n

b) For the another fission reaction:

^{235}_{92}\textrm{U}+^1_0\textrm{n}\rightarrow ^{A}_{56}\textrm{Ba}+^{94}_{Z}\textrm{X}+2^1_0\textrm{n}

To calculate A:

Total mass on reactant side = total mass on product side

235 + 1 = A + 94 + 2

A = 140

To calculate Z:

Total atomic number on reactant side = total atomic number on product side

92 + 0 = 56 + Z + 0

Z = 36

As Krypton has atomic number of 36,Thus the nuclear equation will be :

^{235}_{92}\textrm{U}+^1_0\textrm{n}\rightarrow ^{140}_{56}\textrm{Ba}+^{94}_{36}\textrm{Kr}+2^1_0\textrm{n}

         

8 0
3 years ago
The amount of energy invested into glycolysis is ____________ atp
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The amount of energy invested into glycolysis is 2 ATP.


3 0
3 years ago
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