Colulate the wavelength of i)a ball moving of 40m/s i i)A proton moving at 40 ms i i i) An election moving at 40m/s

The net horizontal force on a car is 981 N. The car has a mass of 1550 kg and the force is applied when the car has a speed of 2

viktelen [127]

Answer:

Distance, d = 778.05 m

Explanation:

Given that,

Force acting on the car, F = 981 N

Mass of the car, m = 1550 kg

Initial speed of the car, v = 25 mi/h = 11.17 m/s

We need to find the distance covered by car if the force continues to be applied to the car. Firstly, lets find the acceleration of the car:

$F=ma\\\\a=\dfrac{F}{m}\\\\a=\dfrac{981}{1550}\\\\a=0.632\ m/s^2$

Let d is the distance covered by car. Using second equation of motion as :

$d=ut+\dfrac{1}{2}at^2\\\\d=11.17\times 35+\dfrac{1}{2}\times 0.632\times (35)^2\\\\d=778.05\ m$

So, the car will cover a distance of 778.05 meters.

5 0

3 years ago

Power plants can be powered by fossil fuels, such as coal or natural gas, or nuclear power to produce steam to turn turbines to

Naddika [18.5K]

Nuclear energy is cleaner while generating electricity. Nuclear fission provides energy without releasing greenhouse gases such as carbon dioxide. However, nuclear power plants generate significant amounts of radioactive waste. That is why we should not choose nuclear energy over fossil fuel power plants.

7 0

2 years ago

A flywheel in a motor is spinning at 590 rpm when a power failure suddenly occurs. The flywheel has mass 40.0 kg and diameter 75

Gnom [1K]

Answer:

Explanation:

Hello,

Let's get the data for this question before proceeding to solve the problems.

Mass of flywheel = 40kg

Speed of flywheel = 590rpm

Diameter = 75cm , radius = diameter/ 2 = 75 / 2 = 37.5cm.

Time = 30s = 0.5 min

During the power off, the flywheel made 230 complete revolutions.

∇θ = [(ω₂ + ω₁) / 2] × t

∇θ = [(590 + ω₂) / 2] × 0.5

But ∇θ = 230 revolutions

∇θ/t = (530 + ω₂) / 2

230 / 0.5 = (530 + ω₂) / 2

Solve for ω₂

460 = 295 + 0.5ω₂

ω₂ = 330rpm

a)

ω₂ = ω₁ + αt

but α = ?

α = (ω₂ - ω₁) / t

α = (330 - 590) / 0.5

α = -260 / 0.5

α = -520rev/min

b)

ω₂ = ω₁ + αt

0 = 590 +(-520)t

520t = 590

solve for t

t = 590 / 520

t = 1.13min

60 seconds = 1min

X seconds = 1.13min

x = (60 × 1.13) / 1

x = 68seconds

∇θ = [(ω₂ + ω₁) / 2] × t

∇θ = [(590 + 0) / 2] × 1.13

∇θ = 333.35 rev/min

8 0

3 years ago

How fast (in rpm) must a centrifuge rotate if a particle 7.3 cm from the axis of rotation is to experience an acceleration of 1.

guajiro [1.7K]

The formula we use here is:

radial acceleration = ω^2 * R 

110,000 * 9.81 m/s^2 = ω^2 * 0.073 m
ω^2 = 110,000 * 9.81 / 0.073
ω = 3844.76 rad/s

and since: ω = 2pi*f --> f = ω/(2pi)
f = 3844.76 / (2pi) = 611.91 rps = 611.91 * 60 rpm
= 36,714.77 rpm

5 0

3 years ago

Place the left charge at the 2 cm position and the right one at the 4 cm position. Vary the left and right charge to the values

Vlad1618 [11]

Answer:

Explanation:

Force between two charges can be expressed as follows

F = k q₁ q₂ / d²

q₁ and q₂ are two charges , d is distance between them , k is a constant whose value is 9 x 10⁹

distance between charges is fixed which is 4 -2 = 2 cm = 2 x 10⁻² m

force between 1μC and 4μC

= 9 x 10⁹ x 1 x 4 x 10⁻¹² / ( 2 x 10⁻² )²

= 9 x 10 = 90 N

force between 4μC and 1μC

= 9 x 10⁹ x 4 x 1 x 10⁻¹² / ( 2 x 10⁻² )²

= 9 x 10 = 90 N

force between 2μC and 2μC

= 9 x 10⁹ x 2 x 2 x 10⁻¹² / ( 2 x 10⁻² )²

= 9 x 10 = 90 N

force between 1μC and 2μC

= 9 x 10⁹ x 1 x 2 x 10⁻¹² / ( 2 x 10⁻² )²

= 4.5 x 10 = 45 N

force between 1μC and 8μC

= 9 x 10⁹ x 1 x 8 x 10⁻¹² / ( 2 x 10⁻² )²

= 18 x 10 = 180 N

force between 2μC and 8μC

= 9 x 10⁹ x 1 x 8 x 10⁻¹² / ( 2 x 10⁻² )²

= 36 x 10 = 360 N

Left Charge Right Charge Resulting force(N)

1μC 4μC 90 N

4μC 1μC 90 N

2μC 2μC 90 N

1μC 2μC 45 N

1μC 8μC 180 N

2μC 8μC 360 N

5 0

2 years ago

Colulate the wavelength of i)a ball moving of 40m/s i i)A proton moving at 40 ms i i i) An election moving at 40m/s​

Colulate the wavelength of i)a ball moving of 40m/s i i)A proton moving at 40 ms i i i) An election moving at 40m/s