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3 years ago

7

A 1.5 kilogram car is moving at 10 meters per second east. A braking force acts on the car for 5.0 seconds, reducing its velocit

y to 2.0 meters second east.
Determine the impulse experienced by the car.

1 answer:

Arada [10]3 years ago

8 0

Answer:

I don't know

Explanation:

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A plate drops onto a smooth floor and shatters into three pieces of equal mass.Two of the pieces go off with equal speeds v at r

Firlakuza [10]

Answer:

Speed of the this part is given as

$v_3 = \sqrt2 v$

Also the direction of the velocity of the third part of plate is moving along 135 degree with respect to one part of the moving plate

Explanation:

As we know by the momentum conservation of the system

we will have

$P_1 + P_2 + P_3 = P_i$

here we know that

$P_1 = P_2$

the momentum of two parts are equal in magnitude but perpendicular to each other

so we will have

$P_1 + P_2 = \sqrt{P^2 + P^2}$

$P_1 + P_2 = \sqrt2 mv$

now from above equation we have

$P_3 = -(P_1 + P_2)$

$mv_3 = -(\sqrt 2 mv)$

$v_3 = \sqrt2 v$

Also the direction of the velocity of the third part of plate is moving along 135 degree with respect to one part of the moving plate

6 0

3 years ago

What do understand by the efficiency of a machine? By using a block and tackel a man can raise a load of 720 N by an effort of 1

motikmotik

Answer:

Efficiency of a machine is how well the machine works and what the machine is capable of doing.

Mechanical advantage=Load/Effort.

720/180=4

6 0

3 years ago

Estimate the wavelength of electrons that have been accelerated from rest through a potential difference of 60 kV.

ivolga24 [154]

Answer: 2.068* $10^{-14}$ m

Explanation: According to work energy-theorem , the workdone in accelerating the electron equals the energy it would give off in terms of light.

workdone= qV

energy = hc/λ

q=magnitude of an electronic charge= 1.602* $10^{-16}$

h= planck constant = 6.626* $10^{-34}$

c= speed of light =2.998* $10^{8}$

v= potential difference= 6* $10^{4}$

λ= wavelength=unknown

by making λ subject of formulae we have that

λ= $\frac{hc}{qv}$

λ = 6.626* $10^{-34}$ * 2.998* $10^{8}$ / 1.602* $10^{-16}$ * 6* $10^{4}$

λ = $\frac{19.878*10^{-26} }{9.612*10^{-12} }$

by doing the necessary calculations, we have that

λ = 2.068* $10^{-14}$ m

8 0

3 years ago

What is the magnitude of the total acceleration of point A after 2 seconds? The bar starts from rest and has a constant angular

monitta

Answer:

a_total = 2 √ (α² + w⁴) , a_total = 2,236 m

Explanation:

The total acceleration of a body, if we use the Pythagorean theorem is

a_total² = a_T²2 + $a_{c}$ ²

where

the centripetal acceleration is

a_{c} = v² / r = w r²

tangential acceleration

a_T = dv / dt

angular and linear acceleration are related

a_T = α r

we substitute in the first equation

a_total = √ [(α r)² + (w r² )²]

a_total = 2 √ (α² + w⁴)

Let's find the angular velocity for t = 2 s if we start from rest wo = 0

w = w₀ + α t

w = 0 + 1.0 2

w = 2.0rad / s

we substitute

a_total = r √(1² + 2²) = r √5

a_total = r 2,236

In order to finish the calculation we need the radius to point A, suppose that this point is at a distance of r = 1 m

a_total = 2,236 m

7 0

3 years ago

A tank contains one mole of carbon dioxide gas at a pressure of 6.70 atm and a temperature of 23.0°C. The tank (which has a fixe

quester [9]

Answer: 888.45 K or 615.3 °c

Explanation:

According to Gay Lussacs law which states that at constant volume, pressure of an ideal gas is directly proportional to it's absolute temperature.

P/T = Constant

Therefore, P1/T1 = P2/T2

P1 = 6.7 atm

T1= 23°c = 273.15 + 23 = 296.15K

Since P2 is tripled, then,

P2 = 6.7 x 3= 20.1 atm

T2 = (20.1 x 296.15) ÷ 6.7

T2 = 888.45 K

Or in celcius 615.3°c

5 0

3 years ago