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2 years ago

The angle between reflected ray and the normal line is

Physics

1 answer:

Alja [10]2 years ago

7 0

Answer:

<h2>Angle(Δ) of Reflection</h2>

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Joey drives along the 110 South freeway and notices a mile marker that reads 260 miles. He drives until he reaches the 150 mile

Nostrana [21]

Answer:

85 miles .

Explanation:

Displacement along the 110 South freeway = 260 - 150 = 110 miles

Displacement along the 110 North freeway = 150 - 175 = - 25 miles

Net displacement = 110 - 25 = 85 miles

So Joey's displacement from the 260 mile marker is 85 miles .

8 0

3 years ago

A 85 kg lineman tackles a 90 kg receiver. The receiver is running 5.8 m/s, and the lineman is moving 4.1 m/s, at a right angle t

weeeeeb [17]

Answer:

3.59 m/s

Explanation:

We are given that

Mass of lineman,m=85 kg

Mass of receiver,m'=90 kg

Speed of receiver,v'=5.8 m/s

Speed of lineman,v=4.1 m/s

$\theta=90^{\circ}$

We have to find the their velocity immediately after the tackle.

Initial momentum, $P_i=\sqrt{p^2_1+p^2_2}=\sqrt{(85\times 4.1)^2+(90\times 5.8)^2}=627.6 kgm/s$

According to law of conservation of momentum

Initial momentum=Final momentum= $(m+m')V$

$627.6=(85+90)V=175V$

$V=\frac{627.6}{175}=3.59 m/s$

3 0

3 years ago

I NEED THIS A SOON AS POSSIBLE

vivado [14]

Work is done when spring is extended or compressed. Elastic potential energy is stored in the spring. Provided inelastic deformation has not happened, the work done is equal to the elastic potential energy stored.

6 0

2 years ago

Read 2 more answers

What is the de Broglie wavelength of an object with a mass of 2.50 kg moving at a speed of 2.70 m/s? (Useful constant: h = 6.63×

xxMikexx [17]

Answer:

9.82 × $10^{-35}$ Hz

Explanation:

De Broglie equation is used to determine the wavelength of a particle (e.g electron) in motion. It is given as:

λ = $\frac{h}{mv}$

where: λ is the required wavelength of the moving electron, h is the Planck's constant, m is the mass of the particle, v is its speed.

Given that: h = 6.63 × $10^{-34}$ Js, m = 2.50 kg, v = 2.70 m/s, the wavelength, λ, can be determined as follows;

λ = $\frac{h}{mv}$

= $\frac{6.63*10^{-34} }{2.5*2.7}$

= $\frac{6.63 * 10^{-34} }{6.75}$

= 9.8222 × $10^{-35}$

The wavelength of the object is 9.82 × $10^{-35}$ Hz.

4 0

3 years ago

A pendulum has 844 J of potential energy at the highest point of its swing. How much kinetic energy will it have at the bottom o

Stolb23 [73]

844J.
Assuming that there were no encumbrances during it's foreswing and it reached it's full potential at apogee.

8 0

3 years ago