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4 years ago

How long does it take for a truck accelerating at 1.5 m/s^2 to got from rest to 75 km/hr

Physics

1 answer:

Free_Kalibri [48]4 years ago

6 0

Answer:

t = 13.9s

Explanation:

u = 0 m/s

v = 75 km/h

= 20.83 m/s

a = 1.5 m/s²

Using

v = u + at

20.83 = 0 + 1.5t

t = 13.9s

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1. Two boxes stacked on top of one another weigh 100 kg total. They apply

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Answer:

150

Explanation:

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3 years ago

Find the uncertainty in position (in nm) associated with an electron that is moving with a velocity of 572 km/s. The uncertainty

Salsk061 [2.6K]

Answer:

Explanation:

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3 years ago

Long, long ago, on a planet far, far away, a physics experiment was carried out. First, a 0.210-kg ball with zero net charge was

tigry1 [53]

Answer:

$\Delta V=316167V$

Explanation:

The difference of electric potential between two points is given by the formula $\Delta V=Ed$ , where d is the distance between them and E the electric field in that region, assuming it's constant.

The electric field formula is $E=\frac{F}{q}$ , where F is the force experimented by a charge q placed in it.

Putting this together we have $\Delta V=\frac{Fd}{q}$ , so we need to obtain the electric force the charged ball is experimenting.

On the second drop, the ball takes more time to reach the ground, this means that the electric force is opposite to its weight W, giving a net force $N=W-F$ . On the first drop only W acts, while on the second drop is N that acts.

Using the equation for accelerated motion (departing from rest) $d=\frac{at^2}{2}$ , so we can get the accelerations for each drop (1 and 2) and relate them to the forces by writting:

$a_1=\frac{2d}{t_1^2}$

$a_2=\frac{2d}{t_2^2}$

These relate with the forces by Newton's 2nd Law:

$W=ma_1$

$N=ma_2$

Putting all together:

$N=W-F=ma_1-F=ma_2$

Which means:

$F=ma_1-ma_2=m(a_1-a_2)=m(\frac{2d}{t_1^2}-\frac{2d}{t_2^2})=2md(\frac{1}{t_1^2}-\frac{1}{t_2^2})$

And finally we substitute:

$\Delta V=\frac{Fd}{q}=\frac{2md^2}{q}(\frac{1}{t_1^2}-\frac{1}{t_2^2})$

Which for our values means:

$\Delta V=\frac{2(0.21Kg)(1m)^2}{7.7\times10^{-6}C}(\frac{1}{(0.35s)^2}-\frac{1}{(0.65s)^2})=316167V$

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3 years ago

This is a type of friction experienced within liquids and gases. It depends on:__________.

Aleksandr-060686 [28]

Answer:

(1) how thick the fluid is viscosity

Explanation:

This is a type of friction experienced within liquids and gases. It depends on:__________.

(1) how thick the fluid is_______?

(2) how the shape of the object?

(3) how the speed of the object?

the thickness of a fluid is known as viscosity. the more viscous a fluid is the more frictional force is exerted on an object by the fluid

frictional force is an opposing force that resist the movement of two surfaces in contact, there are to types 0f frictional force

1. static frictional force

2. dynamic frictional force

4 0

3 years ago

A projectile is launched straight up from a height of 960 feet with an initial velocity of 64 ft/sec. Its height at time t is h(

Natasha2012 [34]

Answer:

a) $t=2s$

b) $h_{max}=1024ft$

c) $v_{y}=-256ft/s$

Explanation:

From the exercise we know the initial velocity of the projectile and its initial height

$v_{y}=64ft/s\\h_{o}=960ft\\g=-32ft/s^2$

To find what time does it take to reach maximum height we need to find how high will it go

b) We can calculate its initial height using the following formula

Knowing that its velocity is zero at its maximum height

$v_{y}^{2}=v_{o}^{2}+2g(y-y_{o})$

$0=(64ft/s)^2-2(32ft/s^2)(y-960ft)$

$y=\frac{-(64ft/s)^2-2(32ft/s^2)(960ft)}{-2(32ft/s^2)}=1024ft$

So, the projectile goes 1024 ft high

a) From the equation of height we calculate how long does it take to reach maximum point

$h=-16t^2+64t+960$

$1024=-16t^2+64t+960$

$0=-16t^2+64t-64$

Solving the quadratic equation

$t=\frac{-b±\sqrt{b^{2}-4ac}}{2a}$

$a=-16\\b=64\\c=-64$

$t=2s$

So, the projectile reach maximum point at t=2s

c) We can calculate the final velocity by using the following formula:

$v_{y}^{2}=v_{o}^{2}+2g(y-y_{o})$

$v_{y}=±\sqrt{(64ft/s)^{2}-2(32ft/s^2)(-960ft)}=±256ft/s$

Since the projectile is going down the velocity at the instant it reaches the ground is:

$v=-256ft/s$

5 0

3 years ago