Calculate the impulse of a 1kg box that starts from rest and accelerates to 50m/s over a period of 10 seconds.

Two copper rods are separated by a small gap at B. Rod AB has a diameter of 200mm and rod BC has a diameter of 150mm. Find the f

gulaghasi [49]

Answer:

A)3.8196 * $10^{9}$ Newtons B) 2.153 * $10^{9}$ Newtons

Explanation:

Force = Pressure * Area.

Pressure is given as 120GPa

Area = Cross Sectional Area of Rod = Area Of A circle = π $R^{2}$ .

Area Expansivity β = $\frac{Change in Area}{Original Area * Temperature Rise}$

Area Expansivity β = 2α

α = 16.9x10-6/ ° C.

Radius of Rod AB = Half 0f Diameter, 200mm = $\frac{0.2m}{2}$ = 0.1 m

Radius of Rod BC = Half Of Diameter, 150mm = $\frac{0.15}{2}$ = 0.075 m.

Note: To convert from millimeter to meter you divide by 1000.

Area of Rod AB = π * $0.1^{2}$ = 0.0314 $m^{2}$

Area of Rod BC = π * $0.075^{2}$ = 0.0177 $m^{2}$ .

Change in Area = 2α * Original Area * Temperature Rise. Therefore for

Rod AB = 2 * 16.9* $10^{-6}$ * 0.0314 * 30 = 3.183 * $10^{-5}$ $m^{2}$ .

For Rod BC we have = 2 * 16.9* $10^{-6}$ * 0.0177 * 30 = 1.794∈-5 $m^{2}$ .

The forces on each rods will be given by.

Force AB = Pressure * Area of Rod AB

= 120GPa * 3.183 * $10^{-5}$ gives 3.8196 * $10^{9}$ Newtons.

Force BC = Pressure * Area of Rod BC

= 120GPa * 1.794 * $10^{-5}$ gives 2.153 * $10^{9}$ Newtons

7 0

3 years ago

A shopper is pushing a cart down a grocery store aisle. Starting from rest, the shopper applies a constant force to the cart for

inysia [295]

A shopping cart that starts from rest, is accelerated for 4 s, moves at constant velocity for 4 s, and is decelerated for 4s until returning to rest, has an average acceleration of 0 m/s².

A shopper is pushing a cart down a grocery store aisle. The movement of the cart is:

It starts from rest.
From t = 0 s to t = 4.0 s it is accelerated with a constant force.
From t = 4 s to t = 8.0 s it receives just enough force to balance the friction on the cart.
From t = 8 s to t = 12 s it is decelerated until it comes to rest.

All in all, at the initial time (t = 0 s), the velocity is 0 m/s (rest) and at the final time (t = 12 s) the velocity is 0 m/s as well (rest). The average acceleration in that period is:

$a = \frac{v_{12}-v__o}{t_{12}-t_0} = \frac{0m/m-0m/s}{12s-0s} = 0 m/s^{2}$

A shopping cart that starts from rest, is accelerated for 4 s, moves at constant velocity for 4 s, and is decelerated for 4s until returning to rest, has an average acceleration of 0 m/s².

Learn more: brainly.com/question/16274121

3 0

3 years ago

Read 2 more answers

One type of slingshot can be made from a length of rope and a leather pocket for holding the stone. The stone can be thrown by w

grigory [225]

Answer:

15.66 rad/s

Explanation:

The vertical motion and horizontal motion are independent of each other.

t = √ ( 2 s/ g) where t = time for the ball to reach the ground and s is the height of the cliff = 18.0 m

t = √ ( 36 / 9.81 ) = 1.916 secs

horizontal distance travel = ut where u is the horizontal velocity of the stone = 30 × r (radius)

tangential velocity V = angular velocity ( ω) × radius

distance traveled = ω × r × t = 30 × r

radius cancelled on both side

ω = 30 / 1.9156 = 15.66 rad/s

4 0

3 years ago

How long does a car (1000 kg) have a speed of 30 m/s from a rest if the engine power is 10kw

Lemur [1.5K]

Answer: 90.1 s

Explanation:

Use equation for power:

P=F*V

Use eqation for force:

F=ma

F---force

V---velocity

Vr=om/s

V=30m/s

m=1000kg

P=10000W

---------------------------

P=FV

F=P/V

F=10000W/30m/s

F=333.33N

Use equation for force to find accelartaion.

F=ma

a=F/m

a=333.33N/1000kg

a=0.333 m/s²

Use equation for accelaration to find out time:

a=(V-Vs)/t

t=(V-Vs)/a

t=(30m/s)/(0.333m/s²)

t=90.09 s≈90.1 s

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5 0

3 years ago

A technique in which the muscles are stretched by an outside force is called _____.

geniusboy [140]

A technique in which the muscles are stretched by an outside force is called Passive Stretching

6 0

3 years ago