Energy that does not involve the large-scale motion or position of objects in a system is called:

A small sandbag is dropped from rest from a hovering hot-air balloon. (assume the positive direction is upward.) (a) after 1.5 s

marishachu [46]

solution:

We know v0 = 0, a = 9.8, t = 4.0. We need to solve for v

so,

we use the equation:

v = v0 + at

v = 0 + 9.8*4.0

v = 39.2 m/s

Now we just need to solve for d, so we use the equation:

d = v0t + 1/2*a*t^2

d = 0*4.0 + 1/2*9.8*4.0^2

d = 78.4 m

3 0

3 years ago

A parachutist falls 50.0 m without friction. When the parachute opens, he slows down at a rate of 67 m/s*2. If he reaches the gr

KIM [24]

Answer:

3.49 seconds

3.75 seconds

-43200 ft/s²

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

$s=ut+\frac{1}{2}at^2\\\Rightarrow 50=0t+\frac{1}{2}\times 9.81\times t^2\\\Rightarrow t=\sqrt{\frac{50\times 2}{9.81}}\\\Rightarrow t=3.19\ s$

Time the parachutist falls without friction is 3.19 seconds

$v^2-u^2=2as\\\Rightarrow v=\sqrt{2as+u^2}\\\Rightarrow v=\sqrt{2\times 9.81\times 50+0^2}\\\Rightarrow v=31.32\ m/s$

Speed of the parachutist when he opens the parachute 31.32 m/s. Now, this will be considered as the initial velocity

$v=u+at\\\Rightarrow 11=31.32+9.81t\\\Rightarrow t=\frac{11-31.32}{-67}=0.3\ s$

So, time the parachutist stayed in the air was 3.19+0.3 = 3.49 seconds

$s=ut+\frac{1}{2}at^2\\\Rightarrow \frac{s}{2}=0t+\frac{1}{2}\times a\times t^2\\\Rightarrow \frac{s}{2}=\frac{1}{2}at^2$

$s=ut+\frac{1}{2}at^2\\\Rightarrow \frac{s}{2}=u1.1+\frac{1}{2}\times a\times 1.1^2$

Now the initial velocity of the last half height will be the final velocity of the first half height.

$v=u+at\\\Rightarrow v=at$

Since the height are equal

$\frac{1}{2}at^2=u1.1+\frac{1}{2}\times a\times 1.1^2\\\Rightarrow \frac{1}{2}at^2=at1.1+\frac{1}{2}\times a\times 1.1^2\\\Rightarrow 0.5t^2-1.1t-0.605=0\\\Rightarrow 500t^2-1100t-605=0$

$t=\frac{11\left(1+\sqrt{2}\right)}{10},\:t=\frac{11\left(1-\sqrt{2}\right)}{10}\\\Rightarrow t=2.65, -0.45$

Time taken to fall the first half is 2.65 seconds

Total time taken to fall is 2.65+1.1 = 3.75 seconds.

When an object is thrown with a velocity upwards then the velocity of the object at the point to where it was thrown becomes equal to the initial velocity.

$v^2-u^2=2as\\\Rightarrow a=\frac{v^2-u^2}{2s}\\\Rightarrow a=\frac{0^2-240^2}{2\times \frac{8}{12}}\\\Rightarrow a=-43200\ ft/s^2$

Magnitude of acceleration is -43200 ft/s²

5 0

3 years ago

A hot brass plate is having its upper surface cooled by impinging jet of air at temperature of 15°c and convection heat transfer

Elodia [21]

200 degrees because I need the points

4 0

3 years ago

A good train took 7 hours to complete its journey. For the first 3 hours, it travelled an average speed of 186km/h. What was the

Sliva [168]

I really hope this helps

4 0

3 years ago

Two identical bowling balls are rolling on a horizontal floor without slipping. The initial speed of both balls is v = 10 m/s. B

Lapatulllka [165]

Answer:

The difference between frictionless ramp and a regular ramp is that on a frictionless ramp the ball cannot roll it can only slide, but on a regular ramp the ball can roll without slipping.

We will use conversation of energy.

$K_A_1 + U_A_1 = K_A_2 + U_A_2\\\frac{1}{2}I\omega^2 + \frac{1}{2}mv^2 + 0 = 0 + mgH_A$

Note that initial potential energy is zero because the ball is on the bottom, and the final kinetic energy is zero because the ball reaches its maximum vertical distance and stops.

For the ball B;

$K_B_1 + U_B_1 = K_B_2 + U_B_2$

$\frac{1}{2}I_B\omega^2 + \frac{1}{2}mv^2 + 0 = 0 + mgH_B$

The initial velocities of the balls are equal. Their maximum climbing point will be proportional to their final potential energy. Since their initial kinetic energies are equal, their final potential energies must be equal as well.

Hence, both balls climb the same point.

Explanation:

4 0

3 years ago