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weqwewe [10]
2 years ago
7

A jet travels 5,400 miles in 9 hours. Its average speed is ____ mph. Type in your numerical answer only; do not type any words o

r letters with your answer.
Physics
1 answer:
KonstantinChe [14]2 years ago
8 0
You divide 5,400 to 9 as speed = d/t
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Sound travels at a rate of 340 m/s in all directions through air. Matt rings a very loud bell at one location, and Steve hears i
Vika [28.1K]

Answer:

110 m/s

Explanation:

because if you subtract 450 from 340 you get 110

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2 years ago
if a sample that is 4860 years old has 50 radium atoms remaining, how many atoms were in the original sample?
snow_lady [41]
243000, take the 4,860 and times it by 50
4 0
3 years ago
An AM radio station broadcasts isotropically (equally in all directions) with an average power of 3.40 kW. A receiving antenna 6
lara [203]

To solve the problem we will apply the concepts related to the Intensity as a function of the power and the area, as well as the electric field as a function of the current, the speed of light and the permeability in free space, as shown below.

The intensity of the wave at the receiver is

I = \frac{P_{avg}}{A}

I = \frac{P_{avg}}{4\pi r^2}

I = \frac{3.4*10^3}{4\pi(4*1609.34)^2} \rightarrow 1mile = 1609.3m

I = 6.529*10^{-6}W/m^2

The amplitude of electric field at the receiver is

I = \frac{E_{max}^2}{2\mu_0 c}

E_{max}= \sqrt{2I\mu_0 c}

The amplitude of induced emf by this signal between the ends of the receiving antenna is

\epsilon_{max} = E_{max} d

\epsilon_{max} = \sqrt{2I \mu_0 cd}

Here,

I = Current

\mu_0 = Permeability at free space

c = Light speed

d = Distance

Replacing,

\epsilon_{max} = \sqrt{2(6.529*10^{-6})(4\pi*10^{-7})(3*10^{8})(60.0*10^{-2})}

\epsilon_{max} = 0.05434V

Thus, the amplitude of induced emf by this signal between the ends of the receiving antenna is 0.0543V

6 0
3 years ago
What is a convex lens?​
Leona [35]

<em>The convex lens is a lens that converges rays of light that convey parallel to its principal axis (i.e. converges the incident rays towards the principal axis) which is relatively thick across the middle and thin at the lower and upper edges. The edges are curved outward rather than inward.</em>

6 0
2 years ago
Read 2 more answers
A wrench 0.500 m long is applied to a nut with a force of 80.0 N. Because of the cramped space, the force must be exerted upward
riadik2000 [5.3K]

Answer:

Torque, \tau=34.6\ N.m

Explanation:

It is given that,

Length of the wrench, l = 0.5 m

Force acting on the wrench, F = 80 N

The force is acting upward at an angle of 60.0° with respect to a line from the bolt through the end of the wrench. We need to find the torque is applied to the nut. We know that torque acting on an object is equal to the cross product of force and distance. It is given by :

\tau=Fr\ sin\theta

\tau=80\times 0.5\ sin(60)

\tau=34.6\ N.m

So, the torque is applied to the nut is 34.6 N.m. Hence, this is the required solution.

7 0
3 years ago
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