A bug starts at point A, crawls 8.0 cm east, then 5.0 cm south, 3.0 west, and 4.0 cm north to point B.

A stone is thrown horizontally from 2.4m above the ground at 35m/s. The wall is 14m away and 1m high.At what height the stone wi

KIM [24]

The stone reaches the wall at a height of 1.62 m.

The stone lands at a point 24.5 m from the point of projection.

The stone is projected horizontally with a velocity u at a height h from the ground. The wall is located at a distance x from the point of projection. The stone takes a time t to reach the wall and in the same time the stone falls a vertical distance y.

The horizontal distance x is traveled with a constant velocity u.

$x=ut$

Calculate the time taken t.

$t=\frac{x}{u} \\ =\frac{14m}{35 m/s} \\ =0.40s$

The stone's initial vertical velocity is zero. It falls through a distance y in the time t under the action of acceleration due to gravity g.

$y=\frac{1}{2} gt^2\\ \frac{1}{2} (9.81m/s^2)(0.40s)^2\\ =0.784m$

The height h₁ of the stone above the ground when it reaches the wall is given by,

$h_1=h-y\\ =(2.4m)-(0.784m)\\ =1.616m=1.62m$

When the stone reaches the wall, its height from the ground is 1.62m.

The stone thus crosses over the wall, since the height of the wall is 1 m. It reaches the ground at a distance R from the point of projection. If the time taken by the stone to reach the ground is t₁, then,

$h=\frac{1}{2} gt_1^2$

Calculate the time taken by the stone to reach the ground.

$t_1=\sqrt{\frac{2h}{g} } \\=\sqrt{\frac{2(2.4m)}{9.81m/s^2} } \\ =0.699 s$

The horizontal distance traveled by the stone is given by,

$R=ut_1 \\ =(35m/s)(0.699s)\\ =24.5m$

The stone lands at point 24.5 m from the point of projection and 10.5 m from the wall.

3 0

1 year ago

Two disks with the same rotational inertia i are spinning about the same frictionless shaft, with the same angular speed ω, but

valentina_108 [34]

Answer:

3. none of these

Explanation:

The rotational kinetic energy of an object is given by:

$K=\frac{1}{2}I \omega^2$

where

I is the moment of inertia

$\omega$ is the angular speed

In this problem, we have two objects rotating, so the total rotational kinetic energy will be the sum of the rotational energies of each object.

For disk 1:

$K_1 = \frac{1}{2}I (\omega)^2 = \frac{1}{2}I\omega^2$

For disk 2:

$K_2 = \frac{1}{2}I(-\omega)^2 = \frac{1}{2}I\omega^2$

so the total energy is

$K=K_1 + K_2 = \frac{1}{2}I\omega^2 + \frac{1}{2}I\omega^2 = I\omega^2$

So, none of the options is correct.

5 0

1 year ago

A person wants to lose weight by "pumping iron". The person lifts an 80 kg weight 1 meter. How many times must this weight be li

statuscvo [17]

Answer:

37357 sec

or 622 min

or 10.4 hrs

Explanation:

GIVEN DATA:

Lifting weight 80 kg

1 cal = 4184 J

from information given in question we have

one lb fat consist of 3500 calories = 3500 x 4184 J

= 14.644 x 10^6 J

Energy burns in 1 lift = m g h

= 80 x 9.8 x 1 = 784 J

lifts required $= \frac{(14.644 x 10^6)}{784}$

= 18679

from the question,

1 lift in 2 sec.

so, total time = 18679 x 2 = 37357 sec

or 622 min

or 10.4 hrs

3 0

1 year ago

The force shown in the attached figure is the net eastward force acting on a ball. The force starts rising at t = 0.012 s, falls

kifflom [539]

Impulse = Integral of F(t) dt from 0.012s to 0.062 s

Given that you do not know the function F(t) you have to make an approximation.

The integral is the area under the curve.

The problem suggest you to approximate the area to a triangle.

In this triangle the base is the time: 0.062 s - 0.012 s = 0.050 s

The height is the peak force: 35 N.

Then, the area is [1/2] (0.05s) (35N) = 0.875 N*s

Answer> 0.875 N*s

6 0

2 years ago

Which description best matches the image below of a hand that is using the right-hand palm rule?

otez555 [7]

Answer:

When reviewing the results, the correct one is C

Explanation:

The right hand rule is widely useful in knowing the direction of force in a maganto field,

The ruler sets the thumb in the direction of the positive particle, the fingers extended in the direction of the magnetic field, and the palm in the direction of the force.

Let's apply this to our exercise.

The thumb that is the speed goes in the negative direction of the axis,

The two extended that the magnetic field look negative x,

The span points entered the dear sheet the negative the Z axis

When reviewing the results, the correct one is C

8 0

2 years ago