All categories

3 years ago

Helllllpppppppppppp.

Physics

1 answer:

JulijaS [17]3 years ago

5 0

Answer:

$\tau=0.03\ N-m$

Explanation:

Given that,

Force acting, F = 6N

The radius of the path, $r=10^{-2}\ m$

Angle, $\theta=30^{\circ}$

We need to find the amount of torque acting on the object. The formula for torque is given by :

$\tau=Fr\sin\theta\\\\\tau=6\times 10^{-2}\times \sin(30)\\\\\tau=0.03\ N-m$

So, the required torque is equal to 0.03 N-m.

You might be interested in

What is the approximate percentage change in the Earth-Sun distance between June 21 and December 21?

tia_tia [17]

For the purpose of the exercise, we can assume that the Earth is at perihelion (closest point to the Sun) on December 21st (in reality, it happens around January 4th) and that the Earth is at aphelion (farthest point from the Sun) on June 21st (in reality, this happens around July 4th). The distance Earth-Sun is the following:
- Perihelion: 147.1 milion km
- Aphelion: 152.1 milion km
- Average distance: 149.6 milion km
So, we can calculate the percentage change with respect to the average distance as:
$d= \frac{d(aphelion)-d(perihelion)}{d(average)}\cdot 100= \frac{152.1-147.1}{149.6}\cdot 100=3.3 \%$

5 0

3 years ago

True or false the law of coservation of charge states that charge is neither created nor destroyed but transferred from one obje

Hoochie [10]

True, charge can not be created or destroyed, only transferred

8 0

4 years ago

Objects falling through air experience a type Of friction called ____

DaniilM [7]

Answer: That's air resistance.

Explanation: Well, air resistance is an upward force exerted on falling objects.

( I hope this helped <3 )

6 0

4 years ago

Read 2 more answers

A fire hose has an inside diameter of 6.40 cm. Suppose such a hose carries a flow of 40.0 L/s starting at a gauge pressure of 1.

inessss [21]

Answer:

The Reynolds numbers for flow in the fire hose.

Explanation:

Given that,

Diameter = 6.40 cm

Rate of flow = 40.0 L/s

Pressure $P=1.62\times10^{6}\ N/m^2$

We need to calculate the Reynolds numbers for flow in the fire hose

Using formula of rate of flow

$Q=Av$

$v=\dfrac{Q}{A}$

Where, Q = rate of flow

A = area of cross section

Put the value into the formula

$v=\dfrac{40.0\times10^{-3}}{3.14\times(3.2\times10^{-2})^2}$

$v=12.44\ m/s$

We need to calculate the Reynolds number

Using formula of the Reynolds number

$n_{R}=\dfrac{2\rho\times v\times r}{\eta}$

Where, $\eta$ =viscosity of fluid

$\rho$ =density of fluid

Put the value into the formula

$n_{R}=\dfrac{2\times100\times12.44\times3.2\times10^{-2}}{1.002\times10^{-3}}$

$n_{R}=7.945\times10^{5}$

Hence, The Reynolds numbers for flow in the fire hose.

3 0

3 years ago

To find the product of 183.5 and 10², move the decimal point in 183.5

S_A_V [24]

Answer:

Given :

To find the product of 183.5 and 10², move the decimal point in 183.5 _____ places to the right because 10² has _____ zeros.

To Find :

We need to fill the spaces.

Solution :

We know, when number with positive power of 10 is multiplied with another number we need to shift the number to the right hand side by the number of zeroes.

So, To find the product of 183.5 and 10², move the decimal point in 183.5 to right places to the right because 10² has two zeros.

Hence, this is the required solution.

Explanation:

please mark me brainliest

5 0

2 years ago