All categories

3 years ago

Give the SI unit. For physics

Physics

1 answer:

IrinaVladis [17]3 years ago

6 0

Answer:

metre (m) - unit of length

kilograms (kg) - unit of mass

second (s) - unit of time

ampere (A) - unit of electrical current

kelvin (K) - unit of temperature

mole (mol) - unit of the amount of substance

Explanation:

You might be interested in

The muscular system and the skeletal system of the human body work together to

AlladinOne [14]

A. Allow movement.
Muscles connect to your skeleton and they contract and move the skeleton along. <span>They help the process of movement happen in a smoother manner.</span>

4 0

3 years ago

Read 2 more answers

How is velocity ratio of wheel and axle calculatedaad<br>

iogann1982 [59]

Answer:

VR = $\frac{Radius of wheel}{Radius of axle}$

Explanation:

A machine is a device that can be used to overcome a load by the application of an effort through a pivot. Examples are: pulleys, wedge, screw jack, wheel and axle etc.

The wheel and axle is a simple device that can be used to lift a load through a height. Its velocity ratio (VR) can be determined by:

VR = $\frac{Radius of wheel}{Radius of axle}$

Note that for a practical wheel and axle, the radius of the wheel is greater than the radius of the axle.

3 0

3 years ago

The spring of a spring balance is 6.0 in. long when there is no weight on the balance, and it is 8.4 in. long with 4.0 lb hung f

Sergeeva-Olga [200]

Answer:

W = 55.12 J

Explanation:

Given,

Natural length = 6 in

Force = 4 lb, stretched length = 8.4 in

We know,

F = k x

k is spring constant

4 = k (8.4-6)

k = 1.67 lb/in

Work done to stretch the spring to 10.1 in.

$W =k\int_{6}^{10.1} x$

$W = \dfrac{k}{2}[x^2]_6^{10.1}$

$W = \dfrac{1}{2}\times 1.67\times (10.1^2-6.0^2)$

W = 55.12 J

Work done in stretching spring from 6 in to 10.1 in is equal to 55.12 J.

3 0

3 years ago

A charged particle is observed traveling in a circular path in a uniform magnetic field. If the particle had been traveling four

8090 [49]

Answer:

The new radius of the trajectory of the particle is four times the previous radius

Explanation:

In order to know what is the radius of the trajectory of the charged particle, if its speed is four times as fast, you take into account the following formula, which describes the radius of a charged particle in a magnetic field:

$r=\frac{mv}{qB}$ (1)

If the speed of the particle is for time as fast, that is, v' = 4v, you obtain, in the equation (1):

$r'=\frac{mv'}{qB}=\frac{m(4v)}{qB}=4\frac{mv}{qB}=4r$

The new radius of the trajectory of the particle is four times the previous radius

8 0

3 years ago

Explain two scenarios where a large truck can have the same momentum as a small car.

KengaRu [80]

The momentum, p, of any object having mass m and the velocity v is

$p=mv\cdots(i)$

Let $M_L$ and $M_S$ be the masses of the large truck and the car respectively, and $V_L$ and V_S be the velocities of the large truck and the car respectively.

So, by using equation (i),

the momentum of the large truck $= M_LV_L$

and the momentum of the small car $= M_SV_S$ .

If the large truck has the same momentum as a small car, then the condition is

$M_LV_L=M_SV_S\cdots(ii)$

The equation (ii) can be rearranged as

$\frac {M_L}{M_S}=\frac {V_S}{V_L} \; or \; \frac{M_L}{V_S}=\frac{M_S}{V_L}$

So, the first scenario:

$\frac {M_L}{M_S}=\frac {V_S}{V_L}$

$\Rhghtarrow M_L:M_S=V_S:V_L$

So, to have the same momentum, the ratio of mass of truck to the mass of the car must be equal to the ratio of velocity of the car to the velocity of the truck.

The other scenario:

$\frac{M_L}{V_S}=\frac{M_S}{V_L}$

$\Rhghtarrow M_L:V_S= M_S:V_L$

So, to have the same momentum, the ratio of mass of truck to the velocity of the car must be equal to the ratio of mass of the car to the velocity of the truck.

5 0

3 years ago