All categories

3 years ago

A butcher has a beam balance and masses 0.5kg and 2kg.How would he measure 1.5kg of meat on the balance at once

Physics

1 answer:

LekaFEV [45]3 years ago

6 0

Answer and Explanation:

The butcher will balance the meat and the 0.5 kg mass on one side of the beam balance and 2 kg mass on the other side of the beam balance.
By so doing, when the beam balances then the amount of meat measured will be exactly 1.5 kg as required. Since the 1.5 kg meat and 0.5 kg mass is equivalent to the 2 kg on the other side of the beam balance.

You might be interested in

A 15.0-kilogram mass is moving at 7.50 meters per second on a horizontal, frictionless surface. What is the

____ [38]

The answer is 570 J. The kinetic energy has the formula of 1/2mV². The total work in this process W= 1/2m(V2²-V1²) = 1/2 * 15.0 * (11.5²-7.50²) = 570 J.

3 0

3 years ago

A student walks 1.0 mi west and then 1.0 mi north. afterward, how far is she from her starting point?

svetlana [45]

1.0 mi because as he has covered 1 the disrance of 1 m and then 1 m in north so he can go straight 1m north so the distance is actually 1m from the starting point

7 0

3 years ago

There are ___________ of galaxies in the universe containing __________ of stars in each galaxy.

aliya0001 [1]

There are Billions and billions of galaxies in the universe containing Trillions and trillions of stars in each galaxy.

8 0

3 years ago

Read 2 more answers

A brick of mass 1.15 kg is attached to a thin (massless) cable and whirled around in a circle in a vertical plane. The circle ha

stira [4]

Explanation:

It is given that,

Mass of the brick, m = 1.15 kg

Radius of the circle, r = 1.44 m

The cable will break if the tension exceeds 43.0 N

Let v is the maximum sped can have at the bottom of the circle before the cable will break. At the bottom of the circle, the net force is equal to the centripetal force along with the weight of the brick. So,

$T=\dfrac{mv^2}{r}+mg$

$\dfrac{T}{m}-g=\dfrac{v^2}{r}$

$v=\sqrt{(\dfrac{T}{m}-g)r}$

$v=\sqrt{(\dfrac{43}{1.15}-9.8)\times 1.44}$

v = 6.30 m/s

So, the maximum speed of the brick at the bottom of the circle before the cable will break is 6.3 m/s. Hence, this is the required solution.

8 0

3 years ago

A person on a cruise ship is doing laps on the promenade deck. On one portion of the track the person is moving north with a spe

Aleksandr-060686 [28]

The direction of motion of the person relative to the water is 16.7° north of east.

Why?

We can solve the problem by applying the Pitagorean Theorem, where the first speed (to the north) and the second speed (to the east) corresponds to two legs of the right triangle formed with them. (north and east directions are perpendicular each other)

We can calculate the angle that give the direction using the following formula:

$Tan(\alpha)=\frac{NorthSpeed}{EastSpeed}\\\\Tan(\alpha)^{-1}=Tan(\frac{NorthSpeed}{EastSpeed})^{-1}\\\\\alpha=Tan(\frac{NorthSpeed}{EastSpeed})^{-1}$

Now, substituting the given information we have:

$alpha=Tan(\frac{NorthSpeed}{EastSpeed})^{-1}$

$\alpha =Tan(\frac{3.6\frac{m}{s} }{12\frac{m}{s} })^{-1}\\\\\alpha =Tan(0.3)^{-1}=16.69\°(North-East)=16.7\°(North-East)$

Hence, we have that the direction of motion of the person relative to the water is 16.7° north of east.

Have a nice day!

7 0

3 years ago