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Nikitich [7]
1 year ago
5

3. Saagar gives a certain teacher some sass followed by an eye roll. Suddenly, a shoe is thrown at him.

Physics
1 answer:
joja [24]1 year ago
8 0

The average acceleration of the shoes thrown at him, a =2.775 m/s².

<h3>Equation :</h3>

To find the acceleration with given data we use this formula,

v = u + aΔt

Where,

v is final velocity

u is initial velocity

a is acceleration

Δt is time

So,

v - u = aΔt

a = v - u /Δt

a = 12km/hr - 0 / 1.2 s

a = 12km/hr / 1.2s

Changing km/hr into m/s,

1 kilometer/hour = 1000meters / 3600 seconds

We get,

12km/hr = 3.33m/s

a = 3.33 m/s / 1.2s

a = 2.775 m/s²

<h3>What is velocity ?</h3>

When an object is moving, its velocity is the rate at which its direction is changing as seen from a specific point of view and as measured by a specific unit of time. Velocity is the rate and direction of an object's movement, whereas speed is the time rate at which an object is moving along a path.

To know more about speed :

brainly.com/question/28224010

#SPJ9

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A force can affect the _____ of an object.
BARSIC [14]
Flow. such as running into something or friction

Flow Final Answer
7 0
3 years ago
Read 2 more answers
What effect does the vertical acceleration have on the horizontal velocity of the projectile?
KengaRu [80]
Answer:
None, if air resistance is ignored.

Explanation:
At any instant, the projectile has vertical and horizontal components of velocity.
Vertical acceleration due to gravity affects the vertical velocity by accelerating the object toward the center of the earth, and by decreasing the upward vertical velocity.. 
The horizontal component of velocity makes the object travel horizontally as long as the projectile is airborne.
Thsi discussion assumes that air resistance is ignored.
3 0
3 years ago
A 100g block lies on an inclined plane that makes an angle of 15 degrees with the horizontal. The coefficient of kinetic frictio
Fed [463]

Answer:

Mass that one should put in the container so that the 100 g block slides down the inclined plane at constant speed = 34.16 g

Explanation:

The vertical forces (with respect to the inclined plane) acting on the 100 g block include the component of the weight of the block in the direction vertical to the inclined plane and the normal reaction of the plane on the block.

And sum of upward forces = sum of downward forces.

N = mg cos θ

m = 100 g = 0.10 kg

g = acceleration due to gravity = 9.8 m/s²

θ = 15°

N = (0.1×9.8×cos 15°) = 0.946582 N

The horizontal forces (With respect to the inclined plane) include the frictional force (acting upwards for the inclined plane, opposite to the intended direction of motion), the Tension in the rope (acting downwards, away from the 100 g block) and the horizontal component (with respect to the inclined plane) of the weight of the block, F, (also acting downards).

For the body to slide down the inclined plane at constant speed, the downward sloping forces must balance the frictional force, that is, there will be no acceleration.

Frictional force = Tension + F

Frictional force = μN

where μ = coefficient of kinetic friction = 0.60

N = normal reaction = 0.9466 N

Frictional force = Fr = (0.60 × 0.9466) = 0.56796 N = 0.568 N

The horizontal component (with respect to the inclined plane) of the weight of the block (also acting downards) = mg sin θ

F = (0.10 × 9.8 × sin 15°) = 0.253624 N

Tension in the rope = T = ?

Fr = F + T

T = Fr - F = 0.568 - 0.253624 = 0.314376 N = 0.3144 N

But the balance on the rope now has the total weight on the container (weight of container + weight on the container) to be equal to 2T.

2T = mg

2 × 0.3144 = 9.8m

m = 0.06416 kg = 64.16 g.

Mass of the container = 30 g

So, mass that one should put in the container so that the 100 g block slides down the inclined plane at constant speed = 64.16 - 30 = 34.16 g

Hope this Helps!!!

8 0
3 years ago
You have a tungsten sphere (emissivity ε = 0.35) of radius 25 cm at a temperature of 25°C. If the sphere is enclosed in a room w
egoroff_w [7]

Answer:

Explanation:

Stefan's formula for emission of radiation is

E = e σ A  ( T⁴ - T₀⁴ )

E is energy radiated , e is emissivity , σ is stefan's constant , T is temperature of object and T₀ is temperature of surrounding. A is area of surface .

E = .35 x 5.67 x 10⁻⁸ ( 298⁴ - 268⁴ ) x 4π x .25²

= 1.9845 x 10⁻⁸ ( 78.86 - 51.58 ) x 10⁸ x .0625

= 3.38  J /s

8 0
2 years ago
X rays of wavelength 0.0169 nm are directed in the positive direction of an x axis onto a target containing loosely bound electr
mamaluj [8]

Answer:

a) 4.04*10^-12m

b) 0.0209nm

c) 0.253MeV

Explanation:

The formula for Compton's scattering is given by:

\Delta \lambda=\lambda_f-\lambda_i=\frac{h}{m_oc}(1-cos\theta)

where h is the Planck's constant, m is the mass of the electron and c is the speed of light.

a) by replacing in the formula you obtain the Compton shift:

\Delta \lambda=\frac{6.62*10^{-34}Js}{(9.1*10^{-31}kg)(3*10^8m/s)}(1-cos132\°)=4.04*10^{-12}m

b) The change in photon energy is given by:

\Delta E=E_f-E_i=h\frac{c}{\lambda_f}-h\frac{c}{\lambda_i}=hc(\frac{1}{\lambda_f}-\frac{1}{\lambda_i})\\\\\lambda_f=4.04*10^{-12}m +\lambda_i=4.04*10^{-12}m+(0.0169*10^{-9}m)=2.09*10^{-11}m=0.0209nm

c) The electron Compton wavelength is 2.43 × 10-12 m. Hence you can use the Broglie's relation to compute the momentum of the electron and then the kinetic energy.

P=\frac{h}{\lambda_e}=\frac{6.62*10^{-34}Js}{2.43*10^{-12}m}=2.72*10^{-22}kgm\\

E_e=\frac{p^2}{2m_e}=\frac{(2.72*10^{-22}kgm)^2}{2(9.1*10^{-31}kg)}=4.06*10^{-14}J\\\\1J=6.242*10^{18}eV\\\\E_e=4.06*10^{-14}(6.242*10^{18}eV)=0.253MeV

5 0
3 years ago
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