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2 years ago

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If the acceleration of the body is towards the center, what is the direction of the unbalanced force ? Using a complete sentence

, describe the direction of the net force that causes the body to travel in a circle at a constant speed.

1 answer:

SOVA2 [1]2 years ago

5 0

Accelerating objects are changing their velocity. Velocity is often thought of as an object's speed with a direction. Thus, objects which are accelerating are either changing their speed or changing their direction. They are either speeding up, slowing down or changing directions. Changing the velocity in any one of these three ways would be an example of an accelerated motion.

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The same physics student jumps off the back of her Laser again, but this time the Laser is

soldi70 [24.7K]

a) The speed of the student after the jump is 1.07 m/s

b) The final speed of the laser is 10.4 m/s

Explanation:

a)

We can solve this problem by applying the law of conservation of momentum: if there are no external forces acting on the system, the total momentum of the student+Laser system must be constant. Therefore, we can write:

$p_i = p_f\\0=mv+MV$

where

The initial momentum is zero

m = 42 kg is the mass of the Laser

v = 1.5 m/s is the final velocity of the Laser

M = 59 kg is the mass of the student

V is the final velocity of the student

Solving the equation for V, we find the velocity of the student:

$V=-\frac{mv}{M}=-\frac{(42)(1.5)}{59}=-1.07 m/s$

So, the final speed of the student is 1.07 m/s.

b)

In this case, the laser and the student are travelling at 3.1 m/s before the student jumps off: therefore, the total momentum before the jump is not zero.

So, the equation of the conservation of momentum is

$(m+M)u=mv+MV$

where

m = 42 kg is the mass of the Laser

M = 59 kg is the student's mass

u = 3.1 m/s is the initial velocity of the student and the Laser

V = -2.1 m/s is the velocity of the student after the jump (she jumps backward)

v is the final velocity of the Laser

And solving for v, we find

$v=\frac{(m+M)u-MV}{m}=\frac{(42+59)(3.1)-(59)(-2.1)}{42}=10.4 m/s$

Learn more about momentum:

brainly.com/question/7973509

brainly.com/question/6573742

brainly.com/question/2370982

brainly.com/question/9484203

#LearnwithBrainly

3 0

2 years ago

Read 2 more answers

What factors affect the ability of a substance to transfer thermal energy to heat or be heated by it's surroundings?

xeze [42]

The prime factors that affect the ability of substances to transfer the thermal energy to heat are the temperature difference between the two objects, area of cross-section, time, and distance travelled by the thermal energy.

<u>Explanation: </u>

The process of heat conduction takes place through contact between two or more objects. But this conduction depends on multiple factors that are responsible for thermal conduction. They are-

Temperature Difference( $\Delta T$ ) - The two objects must have a temperature difference else there will be no thermal conduction between them. The more the difference in their temperatures, the more thermal energy flows from one object to the other.
Area of Cross-section (A) - Larger areas of contact provide as better medium of thermal conduction.
Time (t) - The more time we give for the thermal conduction, the more energy is transmitted from one system to the other.
Distance Travelled (l) - The longer the distance, lesser the conduction. Means, the distance should be minimized in order to achieve the optimum thermal conduction between two objects.

Consider metal pot and its handle, it is being boiled for 15 m. The molecules present near the source of heat, showing fast vibration and bounce off. It actually indicates the heats of substance. That’s why, handle remains hot as heat conduction takes place. It can be estimated by,

$Q=\frac{k A \Delta T t}{l}$

k - Thermal conductivity of the material, measured in J/s.m. $^{\circ} \mathrm{C}$

4 0

2 years ago

A charge of 5.67 x 10-18 C is placed 3.5 x 10 m away from another charge of - 3.79 x 10 "C

miskamm [114]

Answer:

1. 579 x 10 ^-22N

Explanation:

F = kq1q2/r^2

= 9.0 x 10^9 x 5.67 x 10^-18 x 3.79 x 10^-18/ (3.5 x 10^-2)^2

= 1. 579 x 10 ^-22N

6 0

2 years ago

A long, thin straight wire with linear charge density λ runs down the center of a thin, hollow metal cylinder of radius R. The c

netineya [11]

Answer:

$E=\frac{\lambda}{2\pi r\epsilon_0}$

Explanation:

We are given that

Linear charge density of wire= $\lambda$

Radius of hollow cylinder=R

Net linear charge density of cylinder= $2\lambda$

We have to find the expression for the magnitude of the electric field strength inside the cylinder r<R

By Gauss theorem

$\oint E.dS=\frac{q}{\epsilon_0}$

$q=\lambda L$

$E(2\pi rL)=\frac{L\lambda}{\epsilon_0}$

Where surface area of cylinder= $2\pi rL$

$E=\frac{\lambda}{2\pi r\epsilon_0}$

8 0

2 years ago

You have a remote-controlled car that has been programmed to have velocity v⃗ =(−3ti^+2t2j^)m/s, where t is in s. At t = 0 s, th

DiKsa [7]

Answer:

The y-component of the car's position vector is 670m/s.

The x-component of the acceleration vector is -3, and the y-component is 40.

Explanation:

The displacement vector of the car with velocity

$\boldsymbol{v}= (-3t\boldsymbol{i}+2t^2\boldsymbol{j})m/s$

is the integral of the velocity.

Integrating $\boldsymbol{v}$ we get the displacement vector $\boldsymbol{d}$ :

$\boldsymbol{d}=(-\dfrac{3}{2}t^2\boldsymbol{i}+\dfrac{2}{3}t^3\boldsymbol{j} )$

Now if the initial position if the car is

$\boldsymbol{r}= (3.0\boldsymbol{i}+2.0\boldsymbol{j})$

then the displacement of the car at time $t$ is

$\boldsymbol{d(t)}= \boldsymbol{r+d}$

$\boxed{\boldsymbol{d(t)}=(-\dfrac{3}{2}t^2+3.0\boldsymbol{i}+\dfrac{2}{3}t^3+2.0\boldsymbol{j} )}$

Now at $t=10s$ , we have

$\boxed{\boldsymbol{d(t)}=(-147\boldsymbol{i}+670\boldsymbol{j} )}m$

The y-component of the car's position vector is 670m/s.

The acceleration vector is the derivative of the velocity vector:

$\boldsymbol{a(t)}=\dfrac{d\boldsymbol{v(t)}}{dt} =(-3\boldsymbol{i}+4t\boldsymbol{j})$

and at $t=10s$ it is

$\boldsymbol{a(t)}=(-3\boldsymbol{i}+40\boldsymbol{j})m/s^2$

The x-component of the acceleration vector is -3, and the y-component is 40.

5 0

2 years ago