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3 years ago

Determine the time needed for the load at b to attain a speed of 10 m/s, starting from rest.

1 answer:

5 0

Unfortunately, I can't give a definite answer because I only know two information: the initial velocity which is 0 because it is at rest, and the final velocity which is 10 m/s. To solve this, I should know one other information to determine time. It could be the constant acceleration. The useful equation to be used is:

a = (vf - v₀)/t

So, assuming that the object was dropped from a certain height, the acceleration is due to gravity which is equal to 9.81 m/s². Then,

9.81 = (10 - 0)/t
t = 1.02 seconds

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Sometimes students are taught that "air expands as it is heated". And sometimes they are taught that "as air expands it cools".

hram777 [196]

Answer:

Both are true under specific circumstances. And are related to Boyle's law. volume and pressure in a gas are inversely proportional.

Explanation:

There is a tendency to entropy in our reality, that is, in particular true and visible with gases, they tend to occupy the whole space where they are confined, when we heat a volume of gas, then the movement of the particles and in consequence the pressure of the gas increases and to compensate this the volume tends to be increased too, according to Boyle's law. And the opposite happens when the volume is increased, then the pressure is relieved and since the particles are further one from each other, then the temperature is lower, and therefore it cools down.

7 0

3 years ago

6. One minute after takeoff, a rocket carrying the space shuttle into outer space reaches a speed of 447 m/s.

Sidana [21]

Answer:

acceleration of the rocket is given as

$a = 7.45 m/s^2$

Explanation:

As we know that rocket starts from rest and then reach to final speed of 447 m/s after t = 1 min

so we have

$v_i = 0$

$v_f = 447 m/s$

$t = 1 min = 60 s$

so we have

$a = \frac{v_f - v_i}{t}$

$a = \frac{447 - 0}{60}$

$a = 7.45 m/s^2$

7 0

2 years ago

A ball starts from rest and undergoes uniform acceleration of 2.50m/s^2. What is the velocity of the ball 4s later?

erica [24]

Explanation:

Given:

v₀ = 0 m/s

a = 2.50 m/s²

t = 4 s

Find: v

v = at + v₀

v = (2.50 m/s²) (4 s) + 0 m/s

v = 10 m/s

4 0

2 years ago

Read 2 more answers

Me ajudem, Por favor!!!!!!

zzz [600]

Answer:

a) $a=4\,\frac{m}{s^2}$

b) $V(t)=4\,t\,+3$

c) $V(1)=7 \,\frac{m}{s} \\$

d) Displacement = 22 m

e) Average speed = 11 m/s

Explanation:

Notice that the acceleration is the derivative of the velocity function, which in this case, being a straight line is constant everywhere, and which can be calculated as:

$slope= \frac{15=3}{3-0} =4\,\frac{m}{s^2}$

Therefore, acceleration is $a=4\,\frac{m}{s^2}$

b) the functional expression for this line of slope 4 that passes through a y-intercept at (0, 3) is given by:

$y=m\,x+b\\V(t)=4\,t\,+3$

c) Since we know the general formula for the velocity, now we can estimate it at any value for 't", for example for the requested t = 1 second:

$V(t)= 4\,t+3\\V(1)=4\,(1)+3\\V(1)=7 \,\frac{m}{s}$

d) The displacement between times t = 1 sec, and t = 3 seconds is given by the area under the velocity curve between these two time values. Since we have a simple trapezoid, we can calculate it directly using geometry and evaluating V(3) (we already know V(1)):

Displacement = $\frac{(7+15)\,2}{2} = 22\,\,m$

e) Recall that the average of a function between two values is the integral (area under the curve) divided by the length of the interval:

Average velocity = $\frac{22}{2} = 11\, \,\frac{m}{s}$

3 0

3 years ago

A box weighing 10 N is 4 m away from the fulcrum. How much force must be applied 1 m away on the opposite end of the fulcrum in

julia-pushkina [17]

Answer: Socratic

Explanation: i’m not sure but you can use Socratic it’s a good app that helps

4 0

3 years ago