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2 years ago

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A force of 20N changes the position of a body. If mass of the body is 2kg, find the acceleration produced in the body.2. A ball

of mass 500g is thrown upwards with a velocity of 15m/s. Calculate its momentum at the highest point.

1 answer:

shepuryov [24]2 years ago

5 0

Explanation:

<em>Hello</em><em> </em><em>there</em><em>!</em><em>!</em><em>!</em>

<em>You</em><em> </em><em>just</em><em> </em><em>need</em><em> </em><em>to</em><em> </em><em>use</em><em> </em><em>simple</em><em> </em><em>formula</em><em> </em><em>for</em><em> </em><em>force</em><em> </em><em>and</em><em> </em><em>momentum</em><em>, </em>

<em>F</em><em>=</em><em> </em><em>m.a</em>

<em>and</em><em> </em><em>momentum</em><em> </em><em>(</em><em>p</em><em>)</em><em>=</em><em> </em><em>m.v</em>

<em>where</em><em> </em><em>m</em><em>=</em><em> </em><em>mass</em>

<em>v</em><em>=</em><em> </em><em>velocity</em><em>.</em>

<em>a</em><em>=</em><em> </em><em>acceleration</em><em> </em><em>.</em>

<em>And</em><em> </em><em>the</em><em> </em><em>solutions</em><em> </em><em>are</em><em> </em><em>in</em><em> </em><em>pictures</em><em>. </em>

<em><u>Hope</u></em><em><u> </u></em><em><u>it helps</u></em><em><u>.</u></em><em><u>.</u></em>

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Nicki rides her bike at a constant speed for 6 km. That part of her ride takes her 1 h. She then rides her bike at a constant sp

Savatey [412]

km x h = km/h

First trial: 6 x 1 = 6km/h

Second trial: 9 x 2 = 18km/h

6 + 18 = <u>24km/h</u> (Total)

Or

6 + 9 = 15 km

2 + 1 = 3h

15 + 3 = 18

15 x 2 = 30

3 x 2 = 6

30 - 6 = <u>24km/h</u>

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2 years ago

When the boy crashes his bumper car into the girl's bumper car, the momentum from his car is transferred to hers. What evidence

Darya [45]

Answer:

Where is the text?

Explanation:

If you refer to the short sentence you wrote as text, I believe the answer is probably the word "crashes" because it shows how the momentum was transferred.

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2 years ago

Read 2 more answers

A professor's office door is 0.99 m wide, 2.2 m high, 4.2 cm thick; has a mass of 27 kg, and pivots on frictionless hinges. A "d

ANEK [815]

Answer:

$I=8.8209\ kg.m^2$

$\alpha=0.6348\ rad.s^{-2}$

Explanation:

Given:

width of door, $w=0.99\ m$

height of the door, $h=2.2\ m$

thickness of the door, $t=4.2\ cm$

mass of the door, $m=27\ kg$

torque on the door, $\tau=5.6\ N.m$

<em>∵Since the thickness of the door is very less as compared to its other dimensions, therefore we treat it as a rectangular sheet.</em>

For a rectangular sheet we have the mass moment of inertia inertia as:

$I=\frac{1}{3} m.w^2$

$I=\frac{1}{3}\times 27\times 0.99^2$

$I=8.8209\ kg.m^2$

We have a relation between mass moment of inertia, torque and angular acceleration as:

$\alpha=\frac{\tau}{I}$

$\alpha=\frac{5.6}{8.8209}$

$\alpha=0.6348\ rad.s^{-2}$

6 0

2 years ago

A car with speed v and an identical car with speed 2v both travel the same circular section of an unbanked road. If the friction

yawa3891 [41]

Answer:

$F'=\dfrac{F}{4}$

Explanation:

Let m is the mass of both cars. The first car is moving with speed v and the other car is moving with speed 2v. The only force acting on both cars is the centripetal force.

For faster car on the road,

$F=\dfrac{mv^2}{r}$

v = 2v

$F=\dfrac{m(2v)^2}{r}$

$F=4\dfrac{m(v)^2}{r}$ ..........(1)

For the slower car on the road,

$F'=\dfrac{mv^2}{r}$ ............(2)

Equation (1) becomes,

$F=4F'$

$F'=\dfrac{F}{4}$

So, the frictional force required to keep the slower car on the road without skidding is one fourth of the faster car.

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2 years ago

A material that has a fracture toughness of 33 MPa.m0.5 is to be made into a large panel that is 2000 mm long by 250 mm wide and

scoray [572]

Answer:

$F_{allow} = 208.15kN$

Explanation:

The word 'nun' for thickness, I will interpret in international units, that is, mm.

We will begin by defining the intensity factor for the steel through the relationship between the safety factor and the fracture resistance of the panel.

The equation is,

$K_{allow} =\frac{K_c}{N}$

We know that $K_c$ is 33Mpa*m^{0.5} and our Safety factor is 2,

$K_{allow} = \frac{33Mpa*m^{0.5}}{2} = 16.5MPa.m^{0.5}$

Now we will need to find the average width of both the crack and the panel, these values are found by multiplying the measured values given by 1/2

<em>For the crack;</em>

$\alpha = 0.5*L_c = 0.5*4mm = 2mm$

<em>For the panel</em>

$\gamma = 0.5*W = 0.5*250mm = 125mm$

To find now the goemetry factor we need to use this equation

$\beta = \sqrt{sec(\frac{\pi\alpha}{2\gamma})}\\\beta = \sqrt{sec(\frac{2\pi}{2*125mm})}\\\beta = 1$

That allow us to determine the allowable nominal stress,

$\sigma_{allow} = \frac{K_{allow}}{\beta \sqrt{\pi\alpha}}$

$\sigma_{allow} = \frac{16.5}{1*\sqrt{2*10^{-3} \pi}}$

\sigma_{allow} = 208.15Mpa

So to get the force we need only to apply the equation of Force, where

$F_{allow}=\sigma_{allow}*L_c*W$

$F_{allow} = 208.15*250*4$

$F_{allow} = 208.15kN$

That is the maximum tensile load before a catastrophic failure.

4 0

2 years ago