Which is an electromagnetic wave? A. light B. seismic C. sound D. water

Please help me with this question

Oksanka [162]

Car at rest:

velocity= 0m/s

Acceleration:

0.2m/s²

Since total time:

3 min = 180s

Formula of acceleration:

acceleration = [final velocity - initial velocity] ÷ [total time]

Velocity at end:

0.2m/s² = [final velocity - 0m/s] ÷ [180s]

0.2m/s² × 180s = [final velocity]

[final velocity] = 36m/s

Distance travelled:

Velocity = displacement(distance) ÷ time

36m/s = displacement(distance) ÷ 180s

displacement(distance) = 36m/s × 180s

displacement(distance) = 6480m

Hey I'm sorry but i do not understand why the answer on your worksheet for distance travelled is 3240m... its half of what my answer is...

6 0

3 years ago

A chemist must dilute 58.0 ML of 69.6 M aqueous silver nitrate (AgNO3) solution until the concentration falls to 5.00 M. He'll d

sdas [7]

1929292929292929393939339

3 0

3 years ago

The vector sum of the forces acting on the beam is zero, and the sum of the moments about the left end of the beam is zero. (a)

Marysya12 [62]

This question is incomplete, the complete question is;

The vector sum of the forces acting on the beam is zero, and the sum of the moments about the left end of the beam is zero.

(a) Determine the forces and and the couple

(b) Determine the sum of the moments about the right end of the beam.

(c) If you represent the 600-N force, the 200-N force, and the 30 N-m couple by a force F acting at the left end of the beam and a couple M, what is F and M?

Answer:

a)

the x-component of the force at A is $A_{x}$ = 0

the y-component of the force at A is $A_{y}$ = 400 N

the couple acting at A is; $M_{A}$ = 146 N-m

b)

the sum of the momentum about the right end of the beam is; ∑ $M_{R}$ = 0

c)

the equivalent force acting at the left end is; F = -400J ( N)

the couple acting at the left end is; M = - 146 N-m

Explanation:

Given that;

The sum of the forces acting on the beam is zero ∑f = 0

Sum of the moments about the left end of the beam is also zero ∑ $M_{L}$ = 0

Vector force acting at A, $F_{A}$ = $A_{x}i$ + $A_{y}j$

Now, From the image, we have;

a)

∑f = 0

$F_{A}$ - 600j + 200j = 0i + 0j

$A_{x}i$ + $A_{y}j$ - 600j + 200j = 0i + 0j

$A_{x}i$ + ( $A_{y}$ - 400)j = 0i + 0j

now by equating i- coefficients'

$A_{x}$ = 0

so, the x-component of the force at A is $A_{x}$ = 0

also by equating j-coefficient

$A_{y}$ - 400 = 0

$A_{y}$ = 400 N

hence, the y-component of the force at A is $A_{y}$ = 400 N

we also have;

∑ $M_{L}$ = 0

$M_{A}$ - ( 30 N-m ) - ( 0.380 m )( 600 N ) + ( 0.560 m )( 200 N ) = 0

$M_{A}$ - 30 N-m - 228 N-m + 112 Nm = 0

$M_{A}$ - 146 N-m = 0

$M_{A}$ = 146 N-m

Therefore, the couple acting at A is; $M_{A}$ = 146 N-m

b)

The sum of the moments about right end of the beam is;

∑ $M_{R}$ = (0.180 m)(600N) - (30 N-m) - ( 0.56 m)( $A_{y}$ ) + $M_{A}$

∑ $M_{R}$ = (108 N-m) - (30 N-m) - ( 0.56 m)(400 N ) + 146 N-m

∑ $M_{R}$ = (108 N-m) - (30 N-m) - ( 224 N-m ) + 146 N-m

∑ $M_{R}$ = 0

Therefore, the sum of the momentum about the right end of the beam is; ∑ $M_{R}$ = 0

c)

The 600-N force, the 200-N force and the 30 N-m couple by a force F which is acting at the left end of the beam and a couple M.

The equivalent force at the left end will be;

F = -600j + 200j (N)

F = -400J ( N)

Therefore, the equivalent force acting at the left end is; F = -400J ( N)

Also couple acting at the left end

M = -(30 N-m) + (0.560 m)( 200N) - ( 0.380 m)( 600 N)

M = -(30 N-m) + (112 N-m) - ( 228 N-m))

M = 112 N-m - 258 N-m

M = - 146 N-m

Therefore, the couple acting at the left end is; M = - 146 N-m

7 0

3 years ago

What change of states of matter is this? A warm wind makes snow on the ground seem to disappear, but no puddles form.

Aleks04 [339]

Very nice ! That's sublimation . The solid snow proceeds directly
to the vapor state, without ever residing in the liquid state.

7 0

3 years ago

Read 2 more answers

A car drives to the east in a time of 4 hours. Then, immediately (not realistic, but just assume this is the case for this probl

diamong [38]

Answer:

Average speed in east direction $speed=\frac{distance}{time}=\frac{28}{4}=7km/hr$

Explanation:

We have given average speed of the entire trip = 5 km/hr

First the car 4 hours then travels 12 km in 4 hours

So total time of the trip = 4+4 = 8 hours

So total distance traveled in the trip $d=speed\times time = 5\times 8=40km$

As the car travel 12 km in 4 hours to the west

So distance traveled in 4 hour in east = 40 -12 = 28 km

Time in east direction = 4 hour

So average speed in east direction $speed=\frac{distance}{time}=\frac{28}{4}=7km/hr$

4 0

3 years ago