All categories

3 years ago

How do you know that waves sent from the sun to earth are not mechanical waves? Explain

1 answer:

3 0

Mechanical waves require matter to travel medium. If there is no substance or matter to travel through, mechanical waves cannot propagate.

Since space is mostly vacuum, there is no medium for waves to travel through: waves sent by the sun (electromagnetic waves) are not mechanical waves

You might be interested in

At the county fair, Chris throws a 0.12kg baseball at a 2.4kg wooden milk bottle, hoping to knock it off its stand and win a pri

viva [34]

Answer:

$v_{f2}$ =6.5% $v_{i1}$

Explanation:

Mass of the ball: $m_{1} =0.12kg$ ]

Initial velocity of the ball: $v_{i1}$

final velocity of the ball: $v_{f1}$ which is -30/100 of $v_{i1}$ = $-0.3v_{i1}$

Mass of the bottle: $m_{2} =2.4kg$

Initial velocity of the bottle: $v_{i2}=0m/s$

final velocity of the bottle: $v_{f2}$ is unknown (to find)

by using conservation momentum, which stated that the initial momentum is equal to the final momentum.

$m_{1} v_{i1} +m_{2} v_{i2} =m_{1} v_{f1} +m_{2} v_{f2}$

so since the bottle is at rest firstly, therefore $v_{i2} =0$

$m_{1} v_{i1} +m_{2} (0) =m_{1} v_{f1} +m_{2} v_{f2}$

$m_{1} v_{i1} =m_{1} v_{f1} +m_{2} v_{f2}$ equation 1

so now substitute $v_{f1}$ into equation 1

$m_{1} v_{i1} =m_{1} (-0.3v_{i1} ) +m_{2} v_{f2}$

$m_{1} v_{i1} = -0.3m_{1}v_{i1} +m_{2} v_{f2}$

collect the like terms

$m_{1} v_{i1} +0.3m_{1}v_{i1} =m_{2} v_{f2}$

$1.3m_{1} v_{i1} =m_{2} v_{f2}$

divide both side by $m_{2}$

$v_{f2}=\frac{1.3m_{1} v_{i1}}{m_{2} }$

Now substitute

$v_{f2} =\frac{1.3*0.12*v_{i1}}{2.4}\\v_{f2} =\frac{0.156v_{i1} }{2.4} \\v_{f2} =0.065v_{i1}$

$v_{f2} =$ 6.5% $v_{i1}$

6 0

3 years ago

Read 2 more answers

The equation T^2=A^3 shows the relationship between a planet’s orbital period, T, and the planet’s mean distance from the sun, A

kobusy [5.1K]

Answer:

D. 2^(3/2)

Explanation:

Given that

T² = A³

Let the mean distance between the sun and planet Y be x

Therefore,

T(Y)² = x³

T(Y) = x^(3/2)

Let the mean distance between the sun and planet X be x/2

Therefore,

T(Y)² = (x/2)³

T(Y) = (x/2)^(3/2)

The factor of increase from planet X to planet Y is:

T(Y) / T(X) = x^(3/2) / (x/2)^(3/2)

T(Y) / T(X) = (2)^(3/2)

3 0

3 years ago

Experiments using "optical tweezers" measure the elasticity of individual DNA molecules. For small enough changes in length, the

GalinKa [24]

Answer:

Spring constant, k = 0.3 N/m

Explanation:

It is given that,

Force acting on DNA molecule, $F=1.5\ nN=1.5\times 10^{-9}\ N$

The molecule got stretched by 5 nm, $x=5\times 10^{-9}\ m$

Let k is the spring constant of that DNA molecule. It can be calculated using the Hooke's law. It says that the force acting on the spring is directly proportional to the distance as :

$F=-kx$

$k=\dfrac{F}{x}$

$k=\dfrac{1.5\times 10^{-9}\ N}{5\times 10^{-9}\ m}$

k = 0.3 N/m

So, the spring constant of the DNA molecule is 0.3 N/m. Hence, this is the required solution.

8 0

3 years ago

A truck hits a brick wall with a force of 120 N. The collision takes 2.0 seconds.

levacccp [35]

Impulse=force*time
impluse=120N*2.0s
impluse=240 Ns

7 0

3 years ago

A man lifts a 120 kg barbell 2 m above the ground . What is the gain in gravitational PE of the barbell?

SpyIntel [72]

Answer:

2,352 Joules

Explanation:

At the ground, the barbell has a classical mechanical energy value of zero. There is no classical kinetic or potential energy for the barbell. The moment the man starts to lift the barbell, he does work on the barbell and transfers kinetic energy to it due to the motion. At its maximum height where the man lifts the barbell to a stop, the kinetic energy is zero because it transformed into gravitational potential energy stored in the gravitational field. Our reference point for potential was defined to be zero at the floor, therefore we can say that the gravitational potential energy at 2 meters is:

$U=mgh=(120kg)(9.8m/s^2)(2m)=2,352J$

7 0

2 years ago