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1 year ago

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A ball is thrown into the air and follows the parabolic path indicated by the dashed line. Use the arrows provided to indicate t

he approximate directions of the velocity and acceleration of the ball at each of the three locations. (Neglect air resistance.)

1 answer:

avanturin [10]1 year ago

5 0

Answer:

Trajectory is the path of a projectile in air. Any tangential to the curve shows it velocity and the slope of the tangential line will be it acceleration.

Explanation:

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In this lab you will use a cart and track to explore various aspects of motion. You will measure and record the time it takes th

Keith_Richards [23]

It is given that by using track and cart we can record the time and the distance travelled and also the speed of the cart can be recorded. With all this data we can solve questions on the laws of motion.

Like using the first law of motion we can determine the force of gravity acting on the cart that has moved a certain distance and the velocity or the speed of card has already been registered and since time is known putting the values in formula would help us calculate the gravitational pull acting on cart.

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1 year ago

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If the surface temperature of that person's skin is 30∘C (that's a little lower than healthy internal body temperature becaus

anastassius [24]

Answer:

$E=477.92\ W.m^{-2}$

Explanation:

Given that:

Absolute temperature of the body, $T=273+30=303\ K$

emissivity of the body, $\epsilon=1$

<u>Using Stefan Boltzmann Law of thermal radiation:</u>

$E=\epsilon. \sigma.T^4$

where:

$\sigma =5.67\times 10^{-8}\ W.m^{-2}.K^{-4}$ (Stefan Boltzmann constant)

Now putting the respective values:

$E=1\times 5.67\times 10^{-8}\times 303^4$

$E=477.92\ W.m^{-2}$

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1 year ago

Imagine you derive the following expression by analyzing the physics of a particular system: v2=v20+2ax. The problem requires so

lisabon 2012 [21]

As per kinematics equation we are given that

$v^2 = v_o^2 + 2ax$

now we are given that

a = 2.55 m/s^2

$v_0 = 21.8 m/s$

$v = 0$

now we need to find x

from above equation we have

$0^2 = 21.8^2 + 2(2.55)x$

$0 = 475.24 + 5.1 x$

$x = 93.2 m$

so it will cover a distance of 93.2 m

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1 year ago

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Consider a point on a bicycle wheel as the wheel makes exactly four complete revolutions about a fixed axis. Compare the linear

dezoksy [38]

Answer:

Explanation:

Wheel completes four revolution.

The linear displacement is zero.

The angular displacement is 4 x 2π = 8π radian.

So, option (c) is correct.

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1 year ago

Consider a spring that does not obey Hooke’s law very faithfully. One end of the spring is fixed. To keep the spring stretched o

IRINA_888 [86]

Answer:

a) $W=-0.0103125\ J$

b) $W=0.0059375\ J$

c) Compressing is easier

Explanation:

Given:

Expression of force:

$F=kx-bx^2+cx^3$

where:

$k=100\ N.m^{-1}$

$b=700\ N.m^{-2}$

$c=12000\ N.m^{-3}$

$x$ when the spring is stretched

$x$ when the spring is compressed

hence,

$F=100x-700x^2+12000x^3$

a)

From the work energy equivalence the work done is equal to the spring potential energy:

here the spring is stretched so, $x=-0.05\ m$

Now,

The spring constant at this instant:

$j=\frac{F}{x}$

$j=\frac{100\times (-0.05)-700\times (-0.05)^2+12000\times (-0.05)^3}{-0.05}$

$j=-8.25\ N.m^{-1}$

Now work done:

$W=\frac{1}{2} j.x^2$

$W=0.5\times -8.25\times (-0.05)^2$

$W=-0.0103125\ J$

b)

When compressing the spring by 0.05 m

we have, $x=0.05\ m$

<u>The spring constant at this instant:</u>

$j=\frac{F}{x}$

$j=\frac{100\times (0.05)-700\times (0.05)^2+12000\times (0.05)^3}{0.05}$

$j=4.75\ N.m^{-1}$

Now work done:

$W=\frac{1}{2} j.x^2$

$W=0.5\times 4.75\times (0.05)^2$

$W=0.0059375\ J$

c)

Since the work done in case of stretching the spring is greater in magnitude than the work done in compressing the spring through the same deflection. So, the compression of the spring is easier than its stretching.

8 0

1 year ago