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lesya [120]
3 years ago
9

A net torque of magnitude 600 Nm is ex-

Physics
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
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