Ask question

Ask question

All categories

3 years ago

5

Suppose you throw a football upward. Is it true or wrong to say that you have not increased the internal energy of the air withi

n the football

1 answer:

Ahat [919]3 years ago

5 0

Wrong because the energy would decrese

You might be interested in

Consider four point charges arranged in a square with sides of length L. Three of the point charges have charge q and one of the

nydimaria [60]

Answer: $F_{net}=\frac{kq^2}{(L)^2}\left [ \frac{1}{2}+\sqrt{2}\right ]$

Explanation:

Given

Three charges of magnitude q is placed at three corners and fourth charge is placed at last corner with -q charge

Force due to the charge placed at diagonally opposite end on -q charge

$F_1=\frac{kq(-q)}{(L\sqrt{2})^2}$

where $L\sqrt{2}=$ Distance between the two charges

$F_1=-\frac{kq^2}{2L^2}$

negative sign indicates that it is an attraction force

Now remaining two charges will apply the same amount of force as they are equally spaced from -q charge

$F_2=\frac{kq(-q)}{(L)^2}$

The magnitude of force by both the charge is same but at an angle of $90^{\circ}$

thus combination of two forces at 2 and 3 will be

$F'=\sqrt{2}\frac{kq^2}{2L^2}$

Now it will add with force due to 1 charge

Thus net force will be

$F_{net}=\frac{kq^2}{(L)^2}\left [ \frac{1}{2}+\sqrt{2}\right ]$

6 0

3 years ago

You have a summer job at a company that developed systems to safely lower large loads down ramps. Your team is investigating a m

Fofino [41]

Answer:

Note that the emf induced is

emf = B d v cos (A)

---> v = emf / [B d cos (A)]

where

B = magnetic field

d = distance of two rails

v = constant speed

A = angle of rails with respect to the horizontal

Also, note that

I = emf/R

where R = resistance of the bar

Thus,

I = B d v cos (A) / R

Thus, the bar experiences a magnetic force of

F(B) = B I d = B^2 d^2 v cos (A) / R, horizontally, up the incline.

Thus, the component of this parallel to the incline is

F(B //) = F(B) cos(A) = B I d = B^2 d^2 v cos^2 (A) / R

As this is equal to the component of the weight parallel to the incline,

B^2 d^2 v cos^2 (A) / R = m g sin (A)

where m = the mass of the bar.

Solving for v,

v = [R m g sin (A) / B^2 d^2 cos^2 (A)] [ANSWER, the constant speed, PART A]

******************************

v = [R m g sin (A) / B^2 d^2 cos^2 (A)]

Plugging in the units,

m/s = [ [ohm * kg * m/s^2] / [T^2 m^2] ]

Note that T = kg / (s * C), and ohm = J * s/C^2

Thus,

m/s = [ [J * s/C^2 * kg * m/s^2] / [(kg / (s * C))^2 m^2] ]

= [ [J * s/C^2 * kg * m/s^2] / [(kg^2 m^2) / (s^2 C^2)]

As J = kg*m^2/s^2, cancelling C^2,,

= [ [kg*m^2/s^2 * s * kg * m/s^2] / [(kg^2 m^2) / (s^2)]

Cancelling kg^2,

= [ [m^2/s^2 * s * m/s^2] / [(m^2) / (s^2)]

Cancelling m^2/s^2,

= [s * m/s^2]

Cancelling s,

=m/s [DONE! WE SHOWED THE UNITS ARE CORRECT! ]

8 0

3 years ago

WILL NAME THE BRAINLIEST! An airplane undergoes the following displacements: It first flies 72 km in a direction of 30° East of

nekit [7.7K]

Answer:

82.1 km

Explanation:

We need to resolve each displacement along two perpendicular directions: the east-west direction (let's label it with x) and the north-south direction (y). Resolving each vector:

$A_x = (72) sin 30^{\circ} =36.0 km\\A_y = (72) cos 30^{\circ} = 62.4 km$

Vector B is 48 km south, so:

$B_x = 0\\B_y = -48$

Finally, vector C:

$C_x = -(100) cos 30^{\circ} =-86.6 km\\C_y = (100) sin 30^{\circ} = 50.0 km$

Now we add the components along each direction:

$R_x = A_x + B_x + C_x = 36.0 + 0 +(-86.6)=-50.6 km\\R_y = A_y+B_y+C_y = 62.4+(-48)+50.0=64.6 km$

So, the resultant (which is the distance in a straight line between the starting point and the final point of the motion) is

$R=\sqrt{R_x^2+R_y^2}=\sqrt{(-50.6)^2+(64.6)^2}=82.1 km$

4 0

3 years ago

Find the mass and center of mass of the solid E with the given density function ρ. E lies under the plane z = 3 + x + y and abov

makvit [3.9K]

Answer:

The mass of the solid is 16 units.

The center of mass of the solid lies at (0.6875, 0.3542, 2.021)

Work:

Density function: ρ(x, y, z) = 8

x-bounds: [0, 1], y-bounds: [0, x], z-bounds: [0, x+y+3]

The mass M of the solid is given by:

M = ∫∫∫ρ(dV) = ∫∫∫ρ(dx)(dy)(dz) = ∫∫∫8(dx)(dy)(dz)

First integrate with respect to z:

∫∫8z(dx)(dy), evaluate z from 0 to x+y+3

= ∫∫[8x+8y+24](dx)(dy)

Then integrate with respect to y:

∫[8xy+4y²+24y]dx, evaluate y from 0 to x

= ∫[8x²+4x²+24x]dx

Finally integrate with respect to x:

[8x³/3+4x³/3+12x²], evaluate x from 0 to 1

= 8/3+4/3+12

= 16

The mass of the solid is 16 units.

Now we have to find the center of mass of the solid which requires calculating the center of mass in the x, y, and z dimensions.

The z-coordinate of the center of mass Z is given by:

Z = (1/M)∫∫∫ρz(dV) = (1/16)∫∫∫8z(dx)(dy)(dz)

<em>Calculate the integral then divide the result by 16.</em>

First integrate with respect to z:

∫∫4z²(dx)(dy), evaluate z from 0 to x+y+3

= ∫∫[4(x+y+3)²](dx)(dy)

= ∫∫[4x²+24x+8xy+4y²+24y+36](dx)(dy)

Then integrate with respect to y:

∫[4x²y+24xy+4xy²+4y³/3+12y²+36y]dx, evaluate y from 0 to x

= ∫[28x³/3+36x²+36x]dx

Finally integrate with respect to x:

[7x⁴/3+12x³+18x²], evaluate x from 0 to 1

= 7/3+12+18

Z = (7/3+12+18)/16 = <u>2.021</u>

The y-coordinate of the center of mass Y is given by:

Y = (1/M)∫∫∫ρy(dV) = (1/16)∫∫∫8y(dx)(dy)(dz)

<em>Calculate the integral then divide the result by 16.</em>

First integrate with respect to z:

∫∫8yz(dx)(dy), evaluate z from 0 to x+y+3

= ∫∫[8xy+8y²+24y](dx)(dy)

Then integrate with respect to y:

∫[4xy²+8y³/3+12y²]dx, evaluate y from 0 to x

= ∫[20x³/3+12x²]dx

Finally integrate with respect to x:

[5x⁴/3+4x³], evaluate x from 0 to 1

= 5/3+4

Y = (5/3+4)/16 = <u>0.3542</u>

<u />

The x-coordinate of the center of mass X is given by:

X = (1/M)∫∫∫ρx(dV) = (1/16)∫∫∫8x(dx)(dy)(dz)

<em>Calculate the integral then divide the result by 16.</em>

First integrate with respect to z:

∫∫8xz(dx)(dy), evaluate z from 0 to x+y+3

= ∫∫[8x²+8xy+24x](dx)(dy)

Then integrate with respect to y:

∫[8x²y+4xy²+24xy]dx, evaluate y from 0 to x

= ∫[12x³+24x²]dx

Finally integrate with respect to x:

[3x⁴+8x³], evaluate x from 0 to 1

= 3+8 = 11

X = 11/16 = <u>0.6875</u>

<u />

The center of mass of the solid lies at (0.6875, 0.3542, 2.021)

4 0

3 years ago

When will he love me

kipiarov [429]

Put a fork under your pillow tonight, and your wish will come true tomorrow.

4 0

2 years ago

Read 2 more answers