Question attached i need answers quick

A force F of magnitude 2x^3 is applied to stop a particle moving with an initial velocity of v0. The particle travels from x=0 t

3241004551 [841]

Answer:

Explanation:

Given that

F=2x³

Work is given as

The range of x is from x=0 to x=D

W=-∫f(x)dx

Then,

W=-∫2x³dx from x=0 to x=D

W=- 2x⁴/4 from x=0 to x=D

W=-2(D⁴/4-0/4)

W=-D⁴/2

W=1/2D⁴

The correct answer is F

5 0

3 years ago

Which planets are visible without a telescope

Annette [7]

The planet MARS is visible without a telescope on many clear nights. The planets JUPITER, MERCURY, VENUS and SATURN are also viewable without the aid of magnification.

3 0

3 years ago

Ellipses have only one focus. Question 14 options: True False

laila [671]

Answer:

False. This is because ellipses have 2 focus points and not only one.

6 0

2 years ago

The moment of inertia of the empty turntable is 1.5 kg?m2. With a constant torque of 2.5 N?m, the turntableperson system takes 3

Tamiku [17]

$6.0 \mathrm{kg} \mathrm{m}^{2}$ is the persons moment of inertia about an axis through her center of mass.

Answer: Option B

<u>Explanation:</u>

Given data are as follows:

moment of inertia of the empty turntable = 1.5

Torque = 2.5 N/m , and

$\text { Angular acceleration of the turntable }=\frac{\text { angular speed }}{\text { time }}=\frac{1}{3}$

Let the persons moment of inertia about an axis through her center of mass= I

So, Now, from the formula of torque,

$\text { Torque }(\tau)=\text { Moment of inertia(I) } \times \text { Angular acceleration(a) }$

$2.5=(1.5+I) \times \frac{1}{3}$

So, from the above equation, we can measure the person’s moment of Inertia (I)

$2.5 \times 3=1.5+I$

$I=7.5-1.5=6.0 \mathrm{kg} m^{2}$

8 0

3 years ago

While taking off from an aircraft carrier, a jet starting from rest accelerates uniformly to a final speed of 40. meters per sec

mariarad [96]

The acceleration of the jet is $11.4 m/s^2$

Explanation:

Since the jet motion is a uniformly accelerated motion (=constant acceleration), we can use the following suvat equation:

$v^2-u^2=2as$

where

v is the final velocity

u is the initial velocity

a is the acceleration

s is the displacement

For the jet in this problem, we have

u = 0 (it starts from rest)

v = 40 m/s (final velocity)

s = 70 m (length of the runway)

Solving for a, we find the acceleration:

$a=\frac{v-u}{t}=\frac{40^2-0}{2(70)}=11.4 m/s^2$

Learn more about acceleration:

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#LearnwithBrainly

5 0

3 years ago