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3 years ago

9

A net torque of magnitude 600 Nm is ex-

1 answer:

vlabodo [156]3 years ago

5 0

A net torque of magnitude is 600

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What is the acceleration of a body moving with uniform velocity?

AfilCa [17]

Acceleration is the rate of change of velocity, a body moving with uniform velocity does not possess acceleration at all i.e. acceleration is zero

3 0

3 years ago

Q7) A box sliding with a velocity of 5 m/s accelerates at 2 m/s^2. How

grigory [225]

Answer:

The box displacement after 6 seconds is 66 meters.

Explanation:

Let suppose that velocity given in statement represents the initial velocity of the box and, likewise, the box accelerates at constant rate. Then, the displacement of the object ( $\Delta s$ ), in meters, can be determined by the following expression:

$\Delta s = v_{o}\cdot t+\frac{1}{2}\cdot a\cdot t^{2}$ (1)

Where:

$v_{o}$ - Initial velocity, in meters per second.

$t$ - Time, in seconds.

$a$ - Acceleration, in meters per square second.

If we know that $v_{o} = 5\,\frac{m}{s}$ , $t = 6\,s$ and $a = 2\,\frac{m}{s^{2}}$ , then the box displacement after 6 seconds is:

$\Delta s = 66\,m$

The box displacement after 6 seconds is 66 meters.

5 0

3 years ago

If a 580 N net force acts on a 40 kg what will the acceleration of the car be

myrzilka [38]

Answer:

I think u have to do 580N times 40KG

7 0

3 years ago

Read 2 more answers

A vertical spring stretches 4.0 cm when a 12-g object is hung from it. The object is replaced with a block of mass 28 g that osc

marin [14]

Answer:

Time period of the osculation will be 0.0671 sec

Explanation:

It is given a vertical spring is stretched by 4 cm

So change in length of the spring x = 4 cm = 0.04 m

Mass which is hung from it m = 12 gram = 0.012 kg

Sprig force will be equal to weight of the mass

So $kx=mg$

$k\times 0.04=0.012\times 9.8$

k = 244.7 N/m

Now new mass is m = 28 gram = 0.028 kg

So time period with new mass will be

$T=2\pi \sqrt{\frac{m}{k}}$

$=2\times 3.14 \sqrt{\frac{0.028}{244.7}}=0.0671sec$

4 0

3 years ago

51. Suppose you measure the terminal voltage of a 3.200-V lithium cell having an internal resistance of 5.00Ω by placing a 1.00-

Tanya [424]

Explanation:

Given that,

Terminal voltage = 3.200 V

Internal resistance $r= 5.00\ \Omega$

(a). We need to calculate the current

Using rule of loop

$E-IR-Ir=0$

$I=\dfrac{E}{R+r}$

Where, E = emf

R = resistance

r = internal resistance

Put the value into the formula

$I=\dfrac{3.200}{1.00\times10^{3}+5.00}$

$I=3.184\times10^{-3}\ A$

(b). We need to calculate the terminal voltage

Using formula of terminal voltage

$V=E-Ir$

Where, V = terminal voltage

I = current

r = internal resistance

Put the value into the formula

$V=3.200-3.184\times10^{-3}\times5.00$

$V=3.18\ V$

(c). We need to calculate the ratio of the terminal voltage of voltmeter equal to emf

$\dfrac{Terminal\ voltage}{emf}=\dfrac{3.18}{3.200 }$

$\dfrac{Terminal\ voltage}{emf}= \dfrac{159}{160}$

Hence, This is the required solution.

5 0

3 years ago