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3 years ago

According to newton's first law of motion, which force is expected to cause a body to accelerate

Physics

2 answers:

Evgen [1.6K]3 years ago

4 0

Unbalanced force is expected to cause a body to accelerate<span>.

Happy studying ^-^</span>

Mariana [72]3 years ago

4 0

Answer: The answer to this question is Unbalanced force.

Explanation: There are two types of forces:

1. Balanced force: When two forces equal in their magnitude act opposite to each other are known as balanced force. The net force is equal to 0.

2. Unbalanced force: When two forces acting opposite to each other are not equal in magnitude results in the unbalanced force. The net force of this is not equal to 0. Direction of force is in the direction of net force.

Now, according to Newton's first law of motion, which says that force is the product of the object's mass and acceleration. The force should have some value and hence it cannot be equal to zero. therefore the force must be Unbalanced force.

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A boy sleds down a hill and onto a frictionless ice- covered lake at 10.0 m/s. In the middle of the lake is a 1000-kg boulder. W

mina [271]

Answer:

The speed of the sled is 9.2 m/s

The speed of the boulder is 0.82 m/s

Solution:

As per the question:

Mass of the boulder, $m_{B} = 1000\ kg$

Mass of the sled, $m_{S} = 2.50\ kg$

Mass of the boy, $m_{b} = 40\ kg$

Initial Velocity, v = 10.0 m/s

Now,

To calculate the speed of both the sled and the boulder after the occurrence of the collision:

m = $m_{b} + m_{S} = 40 + 2.50 = 42.50\ kg$

Initial velocity of the boulder, $v_{B} = 0\ m/s$

Since, the collision is elastic, both the energy and momentum rem,ain conserved.

Now,

Using the conservation of momentum:

$mv + m_{B}v_{B} = mv' + m_{B}v'_{B}$

where

v' = final velocity of the the system of boy and sled

$v'_{B}$ = final velocity of the boulder

$42.50\times 10 + m_{B}.0 = 42.50v' + 1000v'_{B}$

$42.50v' + 1000v'_{B} = 425$ (1)

Now,

Using conservation of energy:

$\frac{1}{2}mv^{2} + \frac{1}{2}m_{B}v_{B}^{2} = \frac{1}{2}mv'^{2} + \frac{1}{2}m_{B}v'_{B}^{2}$

$42.50\times 10^{2} + m_{B}.0 = 42.50v'^{2} + 1000v'_{B}^{2}$

$42.50v'^{2} + 1000v'_{B}^{2} = 4250$ (2)

Now, from eqn (1) and (2):

$v' = \frac{m - m_{B}}{m + m_{B}}\times v$

$v' = \frac{42.50 - 1000}{42.5 + 1000}\times 10 = - 9.2\ m/s$

Now,

$v'_{B} = \frac{2m}{m + m_{B}}\times v$

$v'_{B} = \frac{2\times 42.50}{42.5 + 1000}\times 10 = 0.82\ m/s$

5 0

3 years ago

Read 2 more answers

when hard stabilization structures such as groins are used to stabilize a shoreline, the change in the longshore current results

Nikitich [7]

When hard stabilization structures such as groins are used to stabilize a shoreline, the change in the longshore current results <u>deposition of sediment. </u>

On the upcurrent side of the barrier, sediment is deposited as the longshore current slows.

What is Hard stabilization?

Hard stabilization is the prevention of erosion through the use of artificial barriers.
Other hard stabilization structures, such as breakwaters and seawalls, are built parallel to the beach to protect the coast from the force of waves.
Hard stabilization structures, such as groins, are built at right angles to the shore to prevent the movement of sand down the coast and maintain the beach.
These constructions are made to last for many years, but because they detract from the visual splendor of the beach, they are not always the ideal answer.
Additionally, they affect the habitats and breeding sites of native shoreline species, interfering with the ecosystem's natural processes.

Learn more about the Hard stabilization with the help of the given link:

brainly.com/question/16022736

#SPJ4

4 0

2 years ago

Bumper car A (281 kg) moving

solmaris [256]

use consevation of linear momentum

m1v1+m2v2=(M1+M2)V3
281(2.82)+209(-1.72)=(209+281)V2
792.42-359.48=490v3
432.9=490v3
V3=0.88m/s

4 0

2 years ago

Derive the dimension of Power

stellarik [79]

Answer:

The dimension of power is energy divided by the time or $[ML^2T^-3]$

Explanation:

Power = $\frac {Work Done}{Time}$

We can derive Dimensions of Power from both formula.

Power = Force * Velocity

As,

Force = mass * acceleration

Therefore, Dimensions of

Force = $[M]*[LT^-2] = [MLT^-2]$

Since,

Velocity = $\frac{Length}{Time}$

Now, Dimension of

Velocity = $[LT^-1]$

We have Both Dimensions,Now we can derive Dimensions Of Power,

Power = Force * Velocity

Power = $[MLT^-2] * [LT^-1]$

Power = $[ML^2T^-3]$

7 0

3 years ago

A 5.50 kg sled is initially at rest on a frictionless horizontal road. The sled is pulled a distance of 3.20 m by a force of 25.

kiruha [24]

(a) 69.3 J

The work done by the applied force is given by:

$W=Fd cos \theta$

where:

F = 25.0 N is the magnitude of the applied force

d = 3.20 m is the displacement of the sled

$\theta=30^{\circ}$ is the angle between the direction of the force and the displacement of the sled

Substituting numbers into the formula, we find

$W=(25.0 N)(3.20 m)(cos 30^{\circ})=69.3 J$

(b) 0

The problem says that the surface is frictionless: this means that no friction is acting on the sled, therefore the energy dissipated by friction must be zero.

(c) 69.3 J

According to the work-energy theorem, the work done by the applied force is equal to the change in kinetic energy of the sled:

$\Delta K = W$

where

$\Delta K$ is the change in kinetic energy

W is the work done

Since we already calculated W in part (a):

W = 69.3 J

We therefore know that the change in kinetic energy of the sled is equal to this value:

$\Delta K=69.3 J$

(d) 4.9 m/s

The change in kinetic energy of the sled can be rewritten as:

$\Delta K=K_f - K_i = \frac{1}{2}mv^2-\frac{1}{2}mu^2$ (1)

where

Kf is the final kinetic energy

Ki is the initial kinetic energy

m = 5.50 kg is the mass of the sled

u = 0 is the initial speed of the sled

v = ? is the final speed of the sled

We can calculate the variation of kinetic energy of the sled, $\Delta K$ , after it has travelled for d=3 m. Using the work-energy theorem again, we find

$\Delta K= W = Fd cos \theta =(25.0 N)(3.0 m)(cos 30^{\circ})=65.0 J$

And substituting into (1) and re-arrangin the equation, we find

$v=\sqrt{\frac{2 \Delta K}{m}}=\sqrt{\frac{2(65.0 J)}{5.50 kg}}=4.9 m/s$

6 0

4 years ago