Is there more than one kind of balance used in a laboratory

The acceleration due to gravity on the surface of Mars is about one-third the acceleration due to gravity on Earthâ€™s surface.

aksik [14]

Answer:

one-third of its weight on Earth's surface

Explanation:

Weight of an object is = W = m*g

Gravity on Earth = g₁ = 9.8 m/s

Gravity on Mars = g₂ = $\frac{1}{3}$ g₁

Weight of probe on earth = w₁ = m * g₁

Weight of probe on Mars = w₂ = m * g₂ -------- ( 1 )

As g₂ = g₁/3 --------- ( 2 )

Put equation (2) in equation (1)

so

Weight of probe on Mars = w₂ = m * g₁ /3

Weight of probe on Mars = $\frac{1}{3}$ m * g₁ = $\frac{1}{3}$ w₁

⇒Weight of probe on Mars = $\frac{1}{3}$ Weight of probe on earth

6 0

4 years ago

PLS ANSWER ASAP!!

Molodets [167]

b) between poles M1 and M2

Explanation:

From the expression, we can deduce that r is the distance between two magnetic poles M1 and M2.

The law of attraction between two magnetic poles states that:

<em> the force of attraction or repulsion between two magnetic poles is a function of the product of the strength of the magnetic poles and the square of the distance between the pole</em>s

Mathematically:

FM = K $\frac{M1 M2}{r^{2} }$

here r is the distance between the poles

FM is the magnetic force between the poles

M1 is the strength of the first magnetic pole

M2 is the strength of the second pole

K is the magnetic field constant

learn more:

magnetic pole brainly.com/question/2191993

#learnwithBrainly

8 0

3 years ago

If a car is traveling on the highway at a constant velocity, the force that pushes the car forward must be A. equal to the weigh

Gala2k [10]

the correct answer is c

4 0

4 years ago

WallyGPX accelerates from 0 m/s to 8 m/s in 3 seconds. What is his acceleration? Is this acceleration higher than that of a car

olga nikolaevna [1]

My Phone is +2348181686682

4 0

4 years ago

The water stream strikes the inclined surface of the cart. Determine the power produced by the stream if, due to rolling frictio

levacccp [35]

Answer

given,

constant speed of cart on right side = 2 m/s

diameter of nozzle = 50 mm = 0.05 m

discharge flow through nozzle = 0.04 m³

One-fourth of the discharge flows down the incline

three-fourths flows up the incline

Power = ?

Normal force i.e. Fn acting on the cart

$F_n = \rho A (v - u)^2 sin \theta$

v is the velocity of jet

Q = A V

$v = \dfrac{0.04}{\dfrac{\pi}{4}d^2}$

$v= \dfrac{0.04}{\dfrac{\pi}{4}\times 0.05^2}$

v = 20.37 m/s

u be the speed of cart assuming it to be u = 2 m/s

angle angle of inclination be 60°

now,

$F_n = 1000 \times \dfrac{\pi}{4}\times 0.05^2\times (20.37 - 2)^2 sin 60^0$

F n = 2295 N

now force along x direction

$F_x = F_n sin 60^0$

$F_x = 2295 \times sin 60^0$

$F_x = 1987.52\ N$

Power of the cart

P = F x v

P = 1987.52 x 20.37

P = 40485 watt

P = 40.5 kW

3 0

3 years ago