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3 years ago

Which mass is undergoing to the greatest amount of acceleration ??

Physics

1 answer:

lina2011 [118]3 years ago

4 0

Answer:

Option (3)

Explanation:

Formula used to calculate acceleration is,

F = ma

Where F = force exerted on a mass

m = mass

a = acceleration due to force exerted on the mass

Option (1),

When F = 100 N and m = 100 kg

100 = 100a

a = 1 m per sec²

Option (2)

For F = 1 N and m = 100 kg

1 = 100a

a = $\frac{1}{100}$

a = 0.01 m per sec²

Option (3)

For F = 100 N and m = 1 kg

100 = 1(a)

a = 100 m per sec²

Option (4)

For F = 1 N and m = 1 kg

1 = 1(a)

a = 1 m per sec²

Therefore. acceleration in Option (3) is the maximum.

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What test were used to find out information about the iceman

Blababa [14]

To find information, the would take DNA tests.

6 0

3 years ago

A police car moving at 36.0 m/s is chasing a speeding motorist traveling at 30.0 m/s. The police car has a siren that emits soun

maxonik [38]

Answer:

The frequency heard by the motorist is 4313.2 Hz.

Explanation:

let f1 be the frequency emited by the police car and f2 be the frequency heard by the motorist, let v1 be the speed of the police car and v2 be the speed of the motorist and v = 343 m/s be the speed of sound.

because the police car is moving towards the motorist at a higher speed, then the motorist will hear a increasing frequency and according to Dopper effect, that frequency is given by:

f1 = [(v + v2/(v - v1))]×(f2)

= [( 343 + 30)/(343 - 36)]×(3550)

= 4313.2 Hz

Therefore, the frequency heard by the motorist is 4313.2 Hz.

5 0

3 years ago

When a transformer is step down then

Vera_Pavlovna [14]

Answer:

Hey!

Your answer should be D!

Explanation:

In a transformer Np / Ns is called the voltage ratio. If Ns is less than Np then Vs is less than Vp. This is called a step-down transformer as the voltage is reduced.

(source from google.com!)

3 0

3 years ago

The time, t, required to drive a fixed distance varies inversely as the speed, r. It takes 3 hr at a speed of 14 km/h to driv

kenny6666 [7]

Answer:

time taken with speed 23 km/h will be 1.8 hours or 1 hour 48 minutes

Explanation:

Given:

Time is inversely proportional to the speed

mathematically,

t ∝ (1/r)

let the proportionality constant be 'k'

thus,

t = k/r

therefore, for case 1

time = 3 hr

speed = 14 km/hr

3 = k/14

also,

for case 2

let the time be = t

r = 23 km/h

thus,

we have

t = k/23

on dividing equation 2 by 1

we get

$\frac{t}{3}=\frac{k/23}{k/14}$

$t=\frac{14\times3}{23}$

t = 1.8 hr = or 1 hour 48 minutes ( 0.8 hours × 60 minutes/hour = 48 minutes)

4 0

3 years ago

One of the waste products of a nuclear reactor is plutonium-239 . This nucleus is radioactive and decays by splitting into a hel

Gekata [30.6K]

Answer:

a) $v_{U-235} = 2.68 \cdot 10^{5} m/s$

$v_{He-4} = -1.57 \cdot 10^{7} m/s$

b) $E_{He-4} = 8.23 \cdot 10^{-13} J$

$E_{U-235} = 1.41 \cdot 10^{-14} J$

Explanation:

Searching the missed information we have:

E: is the energy emitted in the plutonium decay = 8.40x10⁻¹³ J

m(⁴He): is the mass of the helium nucleus = 6.68x10⁻²⁷ kg

m(²³⁵U): is the mass of the helium U-235 nucleus = 3.92x10⁻²⁵ kg

a) We can find the velocities of the two nuclei by conservation of linear momentum and kinetic energy:

Linear momentum:

$p_{i} = p_{f}$

$m_{Pu-239}v_{Pu-239} = m_{He-4}v_{He-4} + m_{U-235}v_{U-235}$

Since the plutonium nucleus is originally at rest, $v_{Pu-239} = 0$ :

$0 = m_{He-4}v_{He-4} + m_{U-235}v_{U-235}$

$v_{He-4} = -\frac{m_{U-235}v_{U-235}}{m_{He-4}}$ (1)

Kinetic Energy:

$E_{Pu-239} = \frac{1}{2}m_{He-4}v_{He-4}^{2} + \frac{1}{2}m_{U-235}v_{U-235}^{2}$

$2*8.40 \cdot 10^{-13} J = m_{He-4}v_{He-4}^{2} + m_{U-235}v_{U-235}^{2}$

$1.68\cdot 10^{-12} J = m_{He-4}v_{He-4}^{2} + m_{U-235}v_{U-235}^{2}$ (2)

By entering equation (1) into (2) we have:

$1.68\cdot 10^{-12} J = m_{He-4}(-\frac{m_{U-235}v_{U-235}}{m_{He-4}})^{2} + m_{U-235}v_{U-235}^{2}$

$1.68\cdot 10^{-12} J = 6.68 \cdot 10^{-27} kg*(-\frac{3.92 \cdot 10^{-25} kg*v_{U-235}}{6.68 \cdot 10^{-27} kg})^{2} +3.92 \cdot 10^{-25} kg*v_{U-235}^{2}$

Solving the above equation for $v_{U-235}$ we have:

$v_{U-235} = 2.68 \cdot 10^{5} m/s$

And by entering that value into equation (1):

$v_{He-4} = -\frac{3.92 \cdot 10^{-25} kg*2.68 \cdot 10^{5} m/s}{6.68 \cdot 10^{-27} kg} = -1.57 \cdot 10^{7} m/s$

The minus sign means that the helium-4 nucleus is moving in the opposite direction to the uranium-235 nucleus.

b) Now, the kinetic energy of each nucleus is:

For He-4:

$E_{He-4} = \frac{1}{2}m_{He-4}v_{He-4}^{2} = \frac{1}{2} 6.68 \cdot 10^{-27} kg*(-1.57 \cdot 10^{7} m/s)^{2} = 8.23 \cdot 10^{-13} J$

For U-235:

$E_{U-235} = \frac{1}{2}m_{U-235}v_{U-235}^{2} = \frac{1}{2} 3.92 \cdot 10^{-25} kg*(2.68 \cdot 10^{5} m/s)^{2} = 1.41 \cdot 10^{-14} J$

I hope it helps you!

3 0

3 years ago