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solniwko [45]
3 years ago
9

PLEASE HELP ME !!!!!

Physics
1 answer:
telo118 [61]3 years ago
4 0

Answer: radiant energy

Explanation:

The energy in electromagnetic waves is sometimes called radiant energy.

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How fast would you have to travel on the surface of earth at the equator to keep up with the sun (that is, so that the sun would
professor190 [17]

Answer: 1037 miles per hour

Explanation: In order to see the sun in the same position in the sky, you would have to travel against the speed of rotation of the earth, because this is what causes the sun to appear in a constantly changing position.

Because of this, we will have to calculate the speed of rotation of the earth. To get started, we must know the circumference of the earth. Assuming the circumference formula for a sphere,

Circumference=2\pi R

Where R is the radius of the earth, we find that the perimeter of the earth is approximately 24881 miles. The equation to calculate speed is given by

v=\frac{Distance}{Time}

Because the earth completes one rotation in 24 hours, we have to find the speed of rotation as the perimeter of the earth divided by 24 hours.

The obtained result is 1037 miles per hour.

You would have to travel at 1037 miles per hour in the direction opposite to the direction the rotation is ocurring in.

5 0
3 years ago
Determine the vector perpendicular to the plane of A= 31+ 6j - 2k and B=4i-j +3k
Sliva [168]

The vector perpendicular to the plane of A = 3i+ 6j - 2k and B = 4i-j +3k is 16 i - 17 j - 27 k

Let r be the vector perpendicular to A and B,

r = A * B

A = 3i + 6j - 2k

B = 4i - j + 3k

a1 = 3

a2 = 6

a3 = - 2

b1 = 4

b2 = - 1

b3 = 3

a * b = ( a2 b3 - b2 a3 ) i + ( a3 b1 - b3 a1 ) j + ( a1 b2 - b1 a2 ) k

a * b = [ ( 6 * 3 ) - ( - 1 * - 2 ) ] i + [ ( - 2 * 4 ) - ( 3 * 3 ) ] j + [ ( 3 * - 1 ) - ( 4 * 6 ) ] k

a * b = 16 i - 17 j - 27 k

The perpendicular vector, r = 16 i - 17 j - 27 k

Therefore, the vector perpendicular to the plane of A = 3i + 6j - 2k and B = 4i - j + 3k is 16 i - 17 j - 27 k

To know more about perpendicular vectors

brainly.com/question/14384780

#SPJ1

5 0
1 year ago
A block lies on a horizontal frictionless surface. A horizontal force of 100 N is applied to the block giving rise to an acceler
KIM [24]

Answer:

(a) m = 33.3 kg

(b) d = 150 m

(c) vf = 30 m/s

Explanation:

Newton's second law to the block:

∑F = m*a Formula (1)

∑F : algebraic sum of the forces in Newton (N)

m : mass s (kg)

a : acceleration  (m/s²)

Data

F= 100 N

a= 3.0 m/s²

(a) Calculating of the  mass of the block:

We replace dta in the formula (1)

F = m*a

100 =  m*3

m = 100 / 3

m = 33.3 kg

Kinematic analysis

Because the block  moves with uniformly accelerated movement we apply the following formulas:

d= v₀t+ (1/2)*a*t² Formula (2)

vf= v₀+a*t   Formula (3)

Where:  

d:displacement in meters (m)  

t : time interval in seconds (s)

v₀: initial speed in m/s  

vf: final speed in m/s  

a: acceleration in m/s²

Data

a= 3.0 m/s²

v₀= 0

t = 10 s

(b) Distance the block will travel if the force is applied for 10 s

We replace dta in the formula (2):

d= v₀t+ (1/2)*a*t²

d = 0+ (1/2)*(3)*(10)²

d =150 m

(c) Calculate the speed of the block after the force has been applied for 10 s

We replace dta in the formula (3):

vf= v₀+a*t

vf= 0+(3*(10)

vf= 30 m/s

4 0
3 years ago
What is the force per unit area at this point acting normal to the surface with unit nor- Side View √√ mal vector n = (1/ 2)ex +
Mumz [18]

Complete Question:

Given \sigma = \left[\begin{array}{ccc}10&12&13\\12&11&15\\13&15&20\end{array}\right] at a point. What is the force per unit area at this point acting normal to the surface with\b n = (1/ \sqrt{2} ) \b e_x + (1/ \sqrt{2}) \b e_z   ? Are there any shear stresses acting on this surface?

Answer:

Force per unit area, \sigma_n = 28 MPa

There are shear stresses acting on the surface since \tau \neq 0

Explanation:

\sigma = \left[\begin{array}{ccc}10&12&13\\12&11&15\\13&15&20\end{array}\right]

equation of the normal, \b n = (1/ \sqrt{2} ) \b e_x + (1/ \sqrt{2}) \b e_z

\b n = \left[\begin{array}{ccc}\frac{1}{\sqrt{2} }\\0\\\frac{1}{\sqrt{2} }\end{array}\right]

Traction vector on n, T_n = \sigma \b n

T_n =  \left[\begin{array}{ccc}10&12&13\\12&11&15\\13&15&20\end{array}\right] \left[\begin{array}{ccc}\frac{1}{\sqrt{2} }\\0\\\frac{1}{\sqrt{2} }\end{array}\right]

T_n = \left[\begin{array}{ccc}\frac{23}{\sqrt{2} }\\0\\\frac{27}{\sqrt{33} }\end{array}\right]

T_n = \frac{23}{\sqrt{2} } \b e_x + \frac{27}{\sqrt{2} } \b e_y + \frac{33}{\sqrt{2} } \b e_z

To get the Force per unit area acting normal to the surface, find the dot product of the traction vector and the normal.

\sigma_n = T_n . \b n

\sigma \b n = (\frac{23}{\sqrt{2} } \b e_x + \frac{27}{\sqrt{2} } \b e_y + \frac{33}{\sqrt{2} } \b e_z) . ((1/ \sqrt{2} ) \b e_x + 0 \b  e_y +(1/ \sqrt{2}) \b e_z)\\\\\sigma \b n = 28 MPa

If the shear stress, \tau, is calculated and it is not equal to zero, this means there are shear stresses.

\tau = T_n  - \sigma_n \b n

\tau =  [\frac{23}{\sqrt{2} } \b e_x + \frac{27}{\sqrt{2} } \b e_y + \frac{33}{\sqrt{2} } \b e_z] - 28( (1/ \sqrt{2} ) \b e_x + (1/ \sqrt{2}) \b e_z)\\\\\tau =  [\frac{23}{\sqrt{2} } \b e_x + \frac{27}{\sqrt{2} } \b e_y + \frac{33}{\sqrt{2} } \b e_z] - [ (28/ \sqrt{2} ) \b e_x + (28/ \sqrt{2}) \b e_z]\\\\\tau =  \frac{-5}{\sqrt{2} } \b e_x + \frac{27}{\sqrt{2} } \b e_y + \frac{5}{\sqrt{2} } \b e_z

\tau = \sqrt{(-5/\sqrt{2})^2  + (27/\sqrt{2})^2 + (5/\sqrt{2})^2} \\\\ \tau = 19.74 MPa

Since \tau \neq 0, there are shear stresses acting on the surface.

3 0
3 years ago
On a straight, level, two-lane road, two cars moving in opposite directions approach and pass each other. Car A is in the eastbo
ludmilkaskok [199]

Answer:

a) 42 m/s, positive direction (to the east), b) 42 m/s, negative direction (to the west).

Explanation:

a) Let consider that Car A is moving at positive direction. Then, the relative velocity of Car A as seen by the driver of Car B is:

\vec v_{A/B} = \vec v_{A} - \vec v_{B}\\\vec v_{A/B} = 11 \frac{m}{s} \cdot i + 31 \frac{m}{s} \cdot i\\\vec v_{A/B} = 42 \frac{m}{s} \cdot i

42 m/s, positive direction (to the east).

b) The relative velocity of Car B as seen by the drive of Car A is:

\vec v_{B/A} = \vec v_{B} - \vec v_{A}\\\vec v_{B/A} = -31 \frac{m}{s} \cdot i - 11 \frac{m}{s} \cdot i\\\vec v_{B/A} = - 42 \frac{m}{s} \cdot i

42 m/s, negative direction (to the west).

5 0
3 years ago
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