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3 years ago

7

A body is moving with a uniform acceleration of 2 m/s2. What does it mean?

2 answers:

kati45 [8]3 years ago

8 0

It means that the body is moving, and getting faster and faster. At any time, it's moving 2 m/s faster than it was moving 1 second earlier.

olchik [2.2K]3 years ago

4 0

Answer:

it means the velocity is changing.

the rate of change of velocity is acceleration

so it means the body is accelerated at 2m/s^2

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What best describes an impulse acting on an object?

Triss [41]

The force on an object use Socratic glad to help ..

8 0

3 years ago

Commercially-available hybrid vehicles, such as the Toyota Prius, use electrical batteries to store energy for later use. Howeve

Bad White [126]

(A) $4.2\cdot 10^5 J$

The energy stored by the system is given by

$E=Pt$

where

P is the power provided

t is the time elapsed

In this case, we have

P = 60 kW = 60,000 W is the power

t = 7 is the time

Therefore, the energy stored by the system is

$E=(60,000 W)(7 s)=4.2\cdot 10^5 J$

(B) 4830 rad/s

The rotational energy of the wheel is given by

$E=\frac{1}{2}I \omega^2$ (1)

where

$I$ is the moment of inertia

$\omega$ is the angular velocity

The moment of inertia of the wheel is

$I=\frac{1}{2}MR^2=\frac{1}{2}(5 kg)(0.12 m)^2=0.036 kg m^2$

where M is the mass and R the radius of the wheel.

We also know that the energy provided is

$E=4.2\cdot 10^5 J$

So we can rearrange eq.(1) to find the angular velocity:

$\omega=\sqrt{\frac{2E}{I}}=\sqrt{\frac{2(4.2\cdot 10^5 J)}{0.036 kg m^2}}=4830 rad/s$

(C) $2.8\cdot 10^6 m/s^2$

The centripetal acceleration of a point on the edge is given by

$a=\omega^2 R$

where

$\omega=4830 rad/s$ is the angular velocity

R = 0.12 m is the radius of the wheel

Substituting, we find

$a=(4830 rad/s)^2 (0.12 m)=2.8\cdot 10^6 m/s^2$

7 0

4 years ago

What is the atomic mass of Jupiter?

lana [24]

The mass of Jupiter is 1.9 x 1027 kg.

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3 years ago

The phenomenon of vehicle "tripping" is investigated here. The sport-utility vehicle is sliding sideways with speed v1 and no an

Mrrafil [7]

Answer:

$v_1 = 3.5 \ m/s$

Explanation:

Given that :

mass of the SUV is = 2140 kg

moment of inertia about G , i.e $I_G$ = 875 kg.m²

We know from the conservation of angular momentum that:

$H_1= H_2$

$mv_1 *0.765 = [I+m(0.765^2+0.895^2)] \omega_2$

$2140v_1*0.765 = [875+2140(0.765^2+0.895^2)] \omega_2$

$1637.1 v_1$ $= 3841.575 \omega_2$

$\omega_2 = \frac{1637.1 v_1}{3841.575}$

$\omega _2 = 0.4626 \ v_1$

From the conservation of energy as well;we have :

$T_2 +V_{2 \to 3} = T_3 \\ \\ \\ \frac{1}{2} I_A \omega_2^2 - mgh =0$

$[\frac{1}{2} [875+2140(0.765^2+0.895^2)](0.4262 \ v_1)^2 -2140(9.81)[\sqrt{0.76^2+0.895^2} -0.765]] =0$

$706.93 \ v_1^2 - 8657.49 =0$

$706.93 \ v_1^2 = 8657.49$

$v_1^2 = \frac{8657.49}{706.93 }$

$v_1 ^2 = 12.25$

$v_1 = \sqrt{ 12.25$

$v_1 = 3.5 \ m/s$

6 0

3 years ago

A wire has a current density of 6.25 × 10 6 A / m 2 6.25×106 A/m2 . If the cross-sectional area of the wire is 1.79 mm 2 1.79 mm

joja [24]

Answer:17.44A

Explanation: Current density=I/Area

Area is given by 2.79mm^2=2.79×10^-6m^2

Current=I=current density ×Area=6.25×10^6 ×2.79×10^-6=17.44A

3 0

3 years ago