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viva [34]
4 years ago
7

A 1.0-kg block and a 2.0-kg block are pressed together on a horizontal frictionless surface with a compressed very light spring

between them. (The blocks are not attached to the spring.) The blocks are then released and the spring uncompresses. Immediately after the blocks lose contact with the spring, which statement below is true?A. the magnitude of the momentum of the heavier block will be greater than the magnitude of the momentum of the lighter block.B. the lighter block will have more kinetic energy than the heavier block.C. both blocks will both have the same amount of kinetic energy.D. both blocks will have equal speeds.E. the heavier block will have more kinetic energy than the lighter block.
Physics
1 answer:
Lunna [17]4 years ago
7 0

Answer:

statement B is true

Explanation:

since same force is applied by the compressed spring on both masses so their rate of change of momenta must be same and since the lighter block has lesser mass so it must have greater velocity to have an equal change in momentum as of heavier mass.

By relation: KE=\frac{(mv)^{2} }{2m}

KE lighter=\frac{(mv)^{2} }{2(1)} , KEheavier=\frac{(mv)^{2} }{2(2)}

comparing momenta of above two equations we get

KElighter (2) = KEheavier (4)

KElighter = 2 KEheavier

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A wildlife researcher is tracking a flock of geese. The geese fly 4.0 km due west, then turn toward the north by 40º and fly ano
Kobotan [32]

Wildlife researcher starts from a and then reaches b, he turns towards north 40 degree to move towards c.

Total displacement is ac

Total horizontal displacement = 4+4 cos40 =7.06 km

Total vertical displacement = 4 sin40 =2.57 km

Total displacement = \sqrt{7.06^2+2.57^2} = 7.51 km

7 0
3 years ago
As part
umka21 [38]

Answer:

Part a)

a = 3.68 m/s^2

Part b)

a = 11.8 m/s^2

Explanation:

Part a)

For force conditions of two blocks we will have

m_1g - T = m_1 a

T - m_2g = m_2 a

now from above equations we have

(m_1 - m_2) g = (m_1 + m_2) a

a = \frac{m_1 - m_2}{m_1 + m_2} g

now we know that

m_1 = \frac{908}{9.8} = 92.65 kg

m_2 = \frac{412}{9.8} = 42 kg

now from above equation we have

a = \frac{92.65 - 42}{92.65 + 42}(9.8)

a = 3.68 m/s^2

Part b)

When heavier block is removed and F = 908 N is applied at the end of the string then we have

F - mg = ma

908 - 412 = 42 a

a = 11.8 m/s^2

8 0
3 years ago
A lava flow is an example of what igneous rock
nydimaria [60]
The answer is extrusive. 
7 0
4 years ago
Read 2 more answers
¿Cuantos metros recorre una motocicleta en un segundo si circula a una velocidad de 90km/h?
Zina [86]

Answer:

La motocicleta recorre 25 metros en 1 segundo si circula a una velocidad de 90 km/h

Explanation:

La velocidad es una magnitud que expresa el desplazamiento que realiza un objeto en una unidad determinada de tiempo, esto es, relaciona el cambio de posición (o desplazamiento) con el tiempo.

Siendo la velocidad es el espacio recorrido en un período de tiempo determinado, entonces 90 km/h indica que en 1 hora la motocicleta recorre 90 km. Entonces, siendo 1 h= 3600 segundos (1 h=60 minutos y 1 minuto=60 segundos) podes aplicar la siguiente regla de tres: si en 3600 segundos (1 hora) la motocicleta recorre 90 km, entonces en 1 segundo ¿cuánta distancia recorrerá?

distancia=\frac{1 segundo*90 km}{3600 segundos}

distancia= 0.025 km

Por otro lado, aplicas la siguiente regla de tres: si 1 km es igual a 1,000 metros, ¿0.025 km cuántos metros son?

distancia=\frac{0.025 km*1,000 metros}{1 km}

distancia= 25 metros

<u><em>La motocicleta recorre 25 metros en 1 segundo si circula a una velocidad de 90 km/h</em></u>

6 0
4 years ago
Can someone please help me with these physics problems? I just don’t even know where to start.
KIM [24]

#1

for the block of mass 5 kg normal force is given as

F_n = mg

F_n = 5*9.8 = 49 N

friction force is given as

F_f = \mu F_n

F_f = 0.1*49 = 4.9 N

Net force is given as

F_{net} = ma

F_{net} = 5*2 = 10 N

now we know that

F_{net} = F_{app} - F_f

10 = F_{app} - 4.9

F_{app} = 14.9 N

#2

Normal force is given as

F_n = mg

F_n = 6*9.8

F_n = 58.8 N

now we know that

F_{net} = F_{app} - F_f

F_{net} = 0

as object moves with constant velocity

F_{app} = F_f = 15 N

now for coefficient of friction we can use

F_f = \mu F_n

15 = \mu * 58.8

\mu = 0.255

#3

net force upwards is given as

F = 1.2 * 10^{-4} N

mass is given as

m = 7 * 10^{-5} kg

now as per newton's law we can say

F = ma

1.2 * 10^{-4} = 7 * 10^{-5} * a

a = 1.71 m/s^2

#4

As we know that when block is sliding on rough surface

part a)

net force = applied force - frictional force

F_{net} = F_{app} - F_f

ma = F_{app} - F_f

5*6 = 40 - F_{f}

F_f = 40 - 30 = 10 N

part b)

for coefficient of friction we can use

F_f = \mu F_n

10 = \mu * F_n

here normal force is given as

F_n = mg = 5*9.8 = 49 N

now we have

\mu = \frac{10}{49} = 0.204

#5

if an object is initially at rest and moves 20 m in 5 s

so we can use kinematics to find out the acceleration

d = v_i*t + \frac{1}{2}at^2

20 = 0 + \frac{1}{2}a(5^2)

a = 1.6 m/s^2

now net force is given as

F_{net} = ma

F_{net} = 10*1.6 = 16 N

#6

an object travelling with speed 25 m/s comes to stop in 1.5 s

so here acceleration of object is given as

a = \frac{v_f - v_i}{t}

a = \frac{0 - 25}{1.5} = -16.67 m/s^2

now the force is gievn as

F = ma

F = 5*16.67 = 83.3 N

3 0
4 years ago
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