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3 years ago

A constant force of 12 N acts for 5 s on a 5 kg object. What is the change in object’s velocity?

Physics

1 answer:

sattari [20]3 years ago

7 0

Answer:

"solve: given that F -12 N and time 4 seconds and let we have to find out the P.

F = 12 N

t = 4 s

p = ?

F = m×( v - u ) / t

12 = m×v / 4

m×v = 12× 4

p = 48 kg m/s

Linear momentum will be 48 kg m/s.

Explanation:

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3 years ago

A 3.00 x 10^2-W electric immersion heater is

andre [41]

Answer

t = 367.77 s = 6.13 min

Explanation:

According to the law of conservation of energy:

$Heat\ Supplied\ By \ Heater = Heat\ Absorbed\ by\ Glass + Heat\ Absorbed\ by\ Water\\Pt = m_gC_g\Delta T_g + m_wC_w\Delta T_w\\$

where,

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t = time required = ?

m_g = mass of glass = 300 g = 0.3 kg

m_w = mass of water = 250 g = 0.25 kg

C_g = speicific heat of glass = 840 J/kg.°C

C_w = specific heatof water = 4184 J/kg.°C

ΔT_g = ΔT_w = Change in Temperature of Glass and water = 100°C - 15°C

ΔT_g = ΔT_w = 85°C

Therefore,

$(300\ W)(t) = (0.3\ kg)(840\ J/kg.^oC)(85^oC)+(0.25\ kg)(4184\ J/kg.^oC)(85^oC)\\$

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3 years ago

a particle of mass m and charge q moving with velocity v in a magnetic field B. The velocity of the particle is perpendicular to

Lerok [7]

Explanation:

When a charged particle describes a circular path in a uniform magnetic field, the charged particle experiences a magnetic force towards the center of circular path, according to Fleming's left hand rule. Therefore the magnetic force and velocity (tangent to circular path) are perpendicular to each other during the circular motion. As the direction of displacement, is the direction of velocity, hence force and displacement are perpendicular to each other.

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3 years ago

An object is projected at 25m/s from the top of a building of height 50m. At the same instant,another object is projected from t

docker41 [41]

A) The objects have the same vertical position after 2 seconds

B) The objects have same vertical position at y = 30.4 m (but they do not collide since they have different x-position)

Explanation:

The motion of the first object along the vertical direction is a uniformly accelerated motion, so we can write its position at time t using the following equation:

$y_1(t)=h+u_1 t + \frac{1}{2}gt^2$

where:

h = 50 m is the initial height

$u_1=0$ is the initial vertical velocity (the object is projected horizontally, so the vertical velocity is zero at the beginning)

$g=-9.8 m/s^2$ is the acceleration of gravity

So, its vertical position can be rewritten as

$y_1(t)=50-4.9t^2$

The position of object 2 instead can be written as

$y_2(t)=(u_2 sin \theta)t + \frac{1}{2}gt^2$

where

$u_2 sin \theta$ is the initial vertical velocity, where

$u_2 = 50 m/s$ is the initial velocity

$\theta=30^{\circ}$ is the angle of projection

Substituting, we get:

$y_2(t)=(50)(sin 30^{\circ})t+\frac{1}{2}(-9.8)t^2=25t-4.9t^2$

The two objects collide when their vertical position is the same, so:

$y_1(t)=y_2(t)\\50-4.9t^2 = 25t-4.9t^2$

And solving for t, we find:

$50=25t\\t= 2 s$

Note that this means that the two object at t = 2 s have the same vertical position: however, this is not true for the horizontal position.

In order to find the point where they collide, we have to substitute the time of the collision that we found in part A into one of the expressions of the vertical position.

Substituting into the expression of object 2, we find:

$y_2(t) = 25t-4.9t^2=25(2.0)-4.9(2.0)^2=30.4 m$