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3 years ago

Which statement is true of gravity?

Physics

1 answer:

o-na [289]3 years ago

3 0

Answer:

B is the answers for the question

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Find the torque required for the shaft to transmit 40 kW when (a) The shaft speed is 2500 rev/min. (b) The shaft speed is 250 re

Luba_88 [7]

Answer:

(a) 152.85 Nm

(b) 1528.5 Nm

Explanation:

According to the formula of power

P = τ ω

ω = 2 π f

(a) f = 2500 rpm = 2500 / 60 = 41.67 rps

So, 40 x 1000 = τ x 2 x 3.14 x 41.67

τ = 152.85 Nm

(b) f = 250 rpm = 250 / 60 = 4.167 rps

So, 40 x 1000 = τ x 2 x 3.14 x 4.167

τ = 1528.5 Nm

3 0

3 years ago

Currents during lightning strikes can be up to 50000 A (or more!). We can model such a strike as a 47500 A vertical current perp

Rufina [12.5K]

Answer:

57 N

Explanation:

Force on a current carrying conductor in a magnetic field

B = 12 X 10⁻⁴ T

= Bil where B is magnetic field , i is current and l is length of conductor

force required = 12 x10⁻⁴ x 47500 x 1

= 57 N

6 0

3 years ago

What graph shape is this what does the snap tell you

vekshin1

The picture is hard to see but if you still need help message me

7 0

3 years ago

An airplane has wings, each with area A, designed so that air flows over the top of the wing at 265 m/s and underneath the wing

exis [7]

Answer

given,

Pressure on the top wing = 265 m/s

speed of underneath wings = 234 m/s

mass of the airplane = 7.2 × 10³ kg

density of air = 1.29 kg/m³

using Bernoulli's equation

$P_1 + \dfrac{1}{2}\rho v_1^2 = P_2 + \dfrac{1}{2}\rho v_2^2$

$\Delta P =\dfrac{1}{2}\rho (v_2^2-v_1^2)$

$\Delta P =\dfrac{1}{2}\times 1.29\times (265^2-234^2)$

$\Delta P =9977.5 Pa$

Applying newtons second law

2 Δ P x A - mg = 0

$A =\dfrac{mg}{2\Delta P}$

$A =\dfrac{7.2\times 10^3 \times 9.8}{2\times 9977.5}$

A = 3.53 m²

7 0

3 years ago

A 1-kg rock is suspended from the tip of a horizontal meterstick at the 0-cm mark so that the meterstick barely balances like a

tigry1 [53]

Explanation:

Given that,

Mass if the rock, m = 1 kg

It is suspended from the tip of a horizontal meter stick at the 0-cm mark so that the meter stick barely balances like a seesaw when its fulcrum is at the 12.5-cm mark.

We need to find the mass of the meter stick. The force acting by the stone is

F = 1 × 9.8 = 9.8 N

Let W be the weight of the meter stick. If the net torque is zero on the stick then the stick does not move and it remains in equilibrium condition. So, taking torque about the pivot.

$9.8\times 12.5=W\times (50-12.5)\\\\W=\dfrac{9.8\times 12.5}{37.5}$

W = 3.266 N

The mass of the meters stick is :

$m=\dfrac{W}{g}\\\\m=\dfrac{3.266}{9.81}\\\\m=0.333\ kg$

So, the mass of the meter stick is 0.333 kg.

5 0

3 years ago