PLEASE HELP!

An umbrella tends to move upward on a windy day because _____.

masha68 [24]

E. all of the above

An umbrella tends to move upward on a windy day because _A. buoyancy increases with increasing wind speed 
B. air gets trapped under the umbrella and pushes it up 
C. the wind pushes it up 
D. a low-pressure area is created on top of the umbrella

3 0

3 years ago

A 2 kg rock is at the edge of a cliff 20 meters above a lake The rock becomes loose and falls toward the water below. Calculate

natima [27]

Answer:

The potential energy (P.E) at the top is 392 J

The kinetic energy (K.E) at the top is 0 J

The potential energy (P.E) at the halfway point is 196 J.

The kinetic energy (K.E) at the halfway point is 196 J.

Explanation:

Given;

mass of the rock, m = 2 kg

height of the cliff, h = 20 m

speed of the rock at the halfway point, v = 14 m/s

The potential energy (P.E) and kinetic energy (K.E) when its at the top;

P.E = mgh

P.E = (2)(9.8)(20)

P.E= 392 J

K.E = ¹/₂mv²

where;

v is velocity of the rock at the top of the cliff = 0

K.E = ¹/₂(2)(0)²

K.E = 0

The potential energy (P.E) and kinetic energy (K.E) at the halfway point;

P.E = mg(¹/₂h)

P.E = (2)(9.8)(¹/₂ x 20)

P.E = 196 J

K.E = ¹/₂mv²

where;

v is velocity of the rock at the halfway point = 14 m/s

K.E = ¹/₂(2)(14)²

K.E = 196 J.

4 0

3 years ago

Let the masses of blocks A and B be 4.50 kg and 2.00 kg , respectively, the moment of inertia of the wheel about its axis be 0.4

Free_Kalibri [48]

Answer:

Accelerations of both the sides is 0.6125 $m/s^{2}$ , A moves downwards whereas B moves upwards.

$\alpha$ =6.125 rad/ $s^{2}$

Tension on side A = 4.5 × g= 44.1 m/ $s^{2}$

Tension on side B= 2.0 × g= 19.6 m/ $s^{2}$

Explanation:

As both, the blocks A and B are attached due to the constraint they can only possess a single acceleration a.

Observe the figure attached, let the tension with Block A be $T_{2}$ and the tension attached with Block B be $T_{1}$ .

Tensions will be only be due to the weight of the blocks as no other force is present.

$T_{2}$ = 4.5 × g= 44.1 m/ $s^{2}$

$T_{1}$ = 2.0 × g= 19.6 m/ $s^{2}$

Now, lets make a torque equation about the center of the wheel and find the alpha

$T_{2}$ ×R- $T_{1}$ ×R= MI( Moment of Inertia of Wheel)× Alpha

where, R= Radius of the wheel=0.100m and

Alpha( $\alpha$ )= Angular acceleration of the wheel

MI of the wheel= 0.400 kg/ $m^{2}$

$(44.1-19.6)R=0.400\alpha$

$\alpha = \frac{24.5 * 0.100}{0.400}$

$\alpha$ =6.125 rad/ $s^{2}$

Acceleration = R × $\alpha$

= 0.1 * 6.125

=0.6125 $m/s^{2}$

Accelerations of both the sides is 0.6125 $m/s^{2}$ , A moves downwards whereas B moves upwards.

7 0

3 years ago

What causes variations in temperature on Earth?

saveliy_v [14]

Earth's temperature depends on the balance between energy entering and leaving the planet's system.

8 0

3 years ago

Read 2 more answers

A ball of mass 0.50 kg is fired with velocity 120 m/s into the barrel of a spring gun of mass 1.6 kg initially at rest on a fric

Gre4nikov [31]

Answer:

The fraction of the ball's kinetic energy stored in the spring is 3.6KJoules

Explanation:

Given

Mass of ball =0.5kg

Velocity of ball =120m/s

The kinetic energy stored by the ball is expressed as

K.E=1/2(m*v²)

Substituting our data into the expression we have

K. E=(0.5*120²)/2

K. E=7200/2

K.E=3,600 JOULE

K.E=3.6KiloJoules

4 0

3 years ago