What force keeps a water spider on the surface of water

A car has the velocity of 2.35 after 4.67 it's velocity is 9.89 what is the average acceleration

kiruha [24]

Average acceleration = (change in speed) / (time for the change) .

Change in speed = (ending speed) - (beginning speed)

=  (9.89 miles/hour) - (2.35 yards/second) = 26,839.2 ft/hr

Acceleration = (26,839.2 ft/hr) / (4.67 days) = 2,873.58 inch/hour²

6 0

3 years ago

1. The gravitational force acting on a falling body and its weight is constant. But the law of universal gravitation tells us th

vagabundo [1.1K]

It's not so much a "contradiction" as an approximation. Newton's law of gravitation is an inverse square law whose range is large. It keeps people on the ground, and it keeps satellites in orbit and that's some thousands of km. The force on someone on the ground - their weight - is probably a lot larger than the centripetal force keeping a satellite in orbit (though I've not actually done a calculation to totally verify this). The distance a falling body - a coin, say - travels is very small, and over such a small distance gravity is assumed/approximated to be constant.

3 0

3 years ago

A coil has 400 turns and self-inductance 7.50 mH. The current in the coil varies with time according to i = 1680 mA2 cos [πt/(0.

zhannawk [14.2K]

Answer:

(a) 1.58 V

(b) 0.0126 Wb

(c) 0.0493 V

Solution:

As per the question:

No. of turns in the coil, N = 400 turns

Self Inductance of the coil, L = 7.50 mH = $7.50\times 10^{- 3}\ H$

Current in the coil, i = $1680cos[\frac{\pi t}{0.0250}]$ A

where

$i_{max} = 1680\ mA = 1.680\ A$

Now,

(a) To calculate the maximum emf:

We know that maximum emf induced in the coil is given by:

$e = \frac{Ldi}{dt}$

$e = L\frac{d}{dt}(1680)cos[\frac{\pi t}{0.0250}]$

$e = - 7.50\times 10^{- 3}\times \frac{\pi}{0.0250}\times \frac{d}{dt}(1680)sin[\frac{\pi t}{0.0250}]$

For maximum emf, $sin\theta$ should be maximum, i.e., 1

Now, the magnitude of the maximum emf is given by:

$|e| = 7.50\times 10^{- 3}\times 1680\times 10^{- 3}\times \frac{\pi}{0.0250} = 1.58\ V$

(b) To calculate the maximum average flux,we know that:

$\phi_{m, avg} = L\times i_{max} = 7.50\times 10^{- 3}\times 1.680 = 0.0126\ Wb$

(c) To calculate the magnitude of the induced emf at t = 0.0180 s:

$e = e_{o}sin{\pi t}{0.0250}$

$e = 7.50\times 10^{- 3}\times sin{\pi \times 0.0180}{0.0250} = 2.96\times 10^{- 4} =0.0493\ V$

7 0

3 years ago

A rock with density 1900 kg/m3 is suspended from the lower end of a light string. When the rock is in air, the tension in the st

wel

Answer:

the tension T2 when the rock is completely immersed is T2 = 29.05 N

Explanation:

from Newton's second law

F= m*a

where F= force , m= mass , a= acceleration

when the rock is suspended ,a=0 since it is at rest. Then

T1 - m*g = 0 , T1= tension when suspended in air , g= gravity

assuming constant density of the rock

m= ρ rock *V , where ρ rock = density of the rock , V= volume

thus

T1= m*g = ρ rock *g*V

V= T1/(ρ rock *g)

when the rock is submerged in oil , it receives an upward force that equals the weight of the volume of displaced oil (V displaced). Since it is completely submerged the volume displaced is the volume of the rock V=Vdisplaced

When the rock is at rest , then

F= m*a=0

T2 + ρ oil *g*V displaced - ρ rock *g*V =0

T2 = ρ rock *g*V - ρ oil *g*V = g*V (ρ rock - ρ oil)

T2 = g*V (ρ rock - ρ oil) = T1/(ρ rock *g) *g * (ρ rock - ρ oil)

T2 = T1 * (ρ rock - ρ oil)/ρ rock

replacing values

T2 = 48 N * (1900 kg/m3- 750 kg/m3)/ 1900 kg/m3 = 29.05 N

T2 = 29.05 N

3 0

3 years ago

Please help me, it's a simple equation

castortr0y [4]

Because the elevator moves at a constant speed, it's in equilibrium and the net force acting on it is zero. Then the tension in the cable exactly equals the magnitude of the elevator's weight, which is

(3000 kg) (9.80 m/s²) = 29,400 N

3 0

2 years ago