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Effectus [21]
2 years ago
10

Two blocks of masses 3.0 kg and 5.0 kg are connected by a spring and rest on a frictionless surface. They are given velocities t

oward each other such that the 3.0 kg block travels initially at 1.2 m/s toward the center of mass, which remains at rest. What is the initial speed of the other block?
Physics
1 answer:
miskamm [114]2 years ago
3 0

Answer:

-0.7 m/sec

Explanation:

Mass of first block = m1 =3.0 kg

Mass of second block = m2= 5.0 kg

Velocity of first block = V1= 1.2 m/s

Velocity of second block = V2 = ?

Momentum of Center of mass MVcom  is sum of both blocks momentum and is given by

MVcom= m1v1+m2v2

Where

M= mass of center of mass

Vcom= Velocity of center of mass=0 m/s (because center of mass is at rest , so Vcom = 0 m.sec)

Putting values, we get;

0= 3×1.2+5v2

==> v2=  3.6/5= - 0.7 m/s

-ve sign indicates that block 2 is moving in opposite direction of block 1

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Vince weighs 160 pounds, and his friend Nadir weighs 140 pounds. Nadir calculated that his weight on another planet would be abo
prisoha [69]

Answer:

C) 64lb

Explanation:

use the linearity method to find the weight of nadir on another planet, it is applied as follows.

Nadir Weight in earth ⇒ Nadir weight in another planet

Vince Weigh in eart  ⇒  X

our goal is to find the weight of vince in another planet (X), for this we multiply the diagonal that continents the data and divide among the remaining

140pounds    ⇒   56lb

160pounds    ⇒ X

X=\frac{(160)(56)}{140} =64lb

Vince weigh on the other planet is C) 64lb

5 0
3 years ago
Read 2 more answers
An electron initially has a speed 16 km/s along the x-direction and enters an electric field of strength 27 mV/m that points in
weqwewe [10]

Answer:

a)t=1.4\times 10^{-5}\ s

b)S= 46.4 cm

Explanation:

Given that

Velocity = 16 Km/s

V= 16,000 m/s

E= 27 mV/m

E=0.027 V/m

d= 22.5 cm

d= 0.225 m

a)

lets time taken by electron is t

d = V x t

0.225 = 16,000 t

t=1.4\times 10^{-5}\ s

b)

We know that

F = m a = E q                    ------------1

Mass of electron ,m

m=9.1\times 10^{-31}\ kg

Charge on electron

q=1.6\times 10^{-19}\ C

So now by putting the values in equation 1

a=\dfrac{E q}{m}

a=\dfrac{1.6\times 10^{-19}\times 0.027}{9.1\times 10^{-31}}\ m/s^2

a=4.74\times 10^{9}\ m/s^2

S= ut+\dfrac{1}{2}at^2

Here initial velocity u= 0 m/s

S= \dfrac{1}{2}\times 4.74\times 10^{9}\times (1.4\times 10^{-5})^2\ m

S=0.464 m

S= 46.4 cm

S is the deflection of electron.

4 0
3 years ago
Read 2 more answers
a hawk flies in a horizontal arc of radius 10.3 m at a constant speed of 4.8 m/s. find its centripetal acceleration. answer in u
n200080 [17]

The hawk’s centripetal acceleration is 2.23 m/s²

The magnitude of the acceleration under new conditions is 2.316 m/s²

radius of the horizontal arc = 10.3 m

the initial constant speed = 4.8 m/s

we know that the centripetal acceleration is given by

    a_{c}  = \frac{v^{2} }{r}

   a_{c}  = 23.04/10.3

    a_{c}  = 2.23 m/s²

It continues to fly but now with some tangential acceleration

a_{t} = 0.63 m/s²

therefore the net value of acceleration is given by the resultant of the centripetal acceleration and the tangential acceleration

so

a_{net}  =  \sqrt{a_{c} ^{2} +a_{t} ^{2}   }

a_{net}  =  \sqrt{4.97 + 0.396}

a_{net}  =  2.316 m/s²

So the magnitude of  net acceleration will become 2.316 m/s².

learn more about acceleration here :

brainly.com/question/11560829

#SPJ4

8 0
1 year ago
Which is an example of a scientist using a conceptual model to describe a volcano?
Yuri [45]
I would say D would be the answer. but is this even a real like homework question 

8 0
3 years ago
Read 2 more answers
The center of a long frictionless rod is pivoted at the origin and the rod is forced to rotate at a constant angular velocity Q
Doss [256]

Answer:

The answers can be found attached.

7 0
2 years ago
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