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Alisiya [41]
3 years ago
12

Which electromagnetic waves has the highest energy?How do you know?

Physics
1 answer:
aliya0001 [1]3 years ago
5 0

Answer:

Gamma Rays

Explanation:

Gamma rays have the highest energies the shortest wavelengths and the highest frequencies. They have the highest energies because they have the highest frequencies, the higher the frequency the more energy. ( I think, maybe double check with other answers )

Good Luck :)

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A 1130-kg car is held in place by a light cable on a smooth (frictionless) ramp. The cable makes an angle of 31.0° above the sur
zubka84 [21]

Answer:

T = 5163.89 N

Explanation:

Newton's first law:

∑F =0 Formula (1)

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

We define the x-axis in the direction parallel to the movement of the car on the ramp and the y-axis in the direction perpendicular to it.

Forces acting on the car

W: Weight of the car : In vertical direction

FN : Normal force : perpendicular to the ramp

T :Tension force:  at angle of 31.0° above the surface of the ramp

Calculated of the Weight  of the car (W)

W = m*g   m: mass   g:acceleration due to gravity

W =   1130-kg* 9.8 m/s² = 11074 N

x-y weight components

Wx =  11074 N*sin 25.0° = 4680.07 N

Wy = 11074 N*cos 25.0° = 10036.45 N

x-y Tension components

Tx = T*cos 25.0°

Ty = T*sin 25.0°

Newton's first law:

∑Fx =0 Formula (1)

Tx-Wx = 0

T*cos 25.0° - 4680.07 = 0

T*cos 25.0° = 4680.07

T =  4680.07 / cos 25.0°

T = 5163.89 N

4 0
3 years ago
A jogger accelerates from rest to 4.86 m/s in 2.43 s. A car accelerates from 20.6 to 32.7 m/s also in 2.43 s. (a) Find the magni
Aleonysh [2.5K]

Explanation:

It is given that,

Initially, the jogger is at rest u₁ = 0

He accelerates from rest to 4.86 m, v₁ = 4.86 m

Time, t₁ = 2.43 s

A car accelerates from u₂ = 20.6 to v₂ = 32.7 m/s in t₂ = 2.43 s

(a) Acceleration of the jogger :

a=\dfrac{v-u}{t}

a=\dfrac{4.86\ m/s-0}{2.43\ s}

a₁ = 2 m/s²

(b) Acceleration of the car,

a=\dfrac{v-u}{t}

a=\dfrac{32.7\ m/s-20.6\ m/s}{2.43\ s}

a₂ = 4.97 m/s²

(c) Distance covered by the car,

d_1=u_1t_1+\dfrac{1}{2}a_1t_1^2

d_1=0+\dfrac{1}{2}\times 2\times (2.43)^2

d₁ = 5.904 m

Distance covered by the jogger,

d_2=u_2t_2+\dfrac{1}{2}a_2t_2^2

d_2=20.6\times 2.43+\dfrac{1}{2}\times 4.97\times (2.43)^2

d₂ = 64.73 m

The car further travel a distance of, d = 64.73 m - 5.904 m = 58.826 m

Hence, this is the required solution.

5 0
2 years ago
PLEASE HALP ;w;
givi [52]
A haha snake e enajene. skskskksksks and
4 0
2 years ago
What are tiny sacs at the end of the bronchioles filled with air called?
WARRIOR [948]

Answer:

Alveoli

The bronchioles end in tiny air sacs called alveoli, where oxygen is transferred from the inhaled air to the blood.

Hope this helps :)

7 0
3 years ago
for an ideal monoatomic gas, the internal energy U os due to the kinetic energy and U=3/2RT per mole.show that cv=3/2R per mole
sladkih [1.3K]

Answer:

i. Cv =3R/2

ii. Cp = 5R/2

Explanation:

i. Cv = Molar heat capacity at constant volume

Since the internal energy of the ideal monoatomic gas is U = 3/2RT and Cv = dU/dT

Differentiating U with respect to T, we have

= d(3/2RT)/dT

= 3R/2

ii. Cp - Molar heat capacity at constant pressure

Cp = Cv + R

substituting Cv into the equation, we have

Cp = 3R/2 + R

taking L.C.M

Cp = (3R + 2R)/2

Cp = 5R/2

3 0
3 years ago
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