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noname [10]
3 years ago
10

Richardson pulls a toy 3.0 m across the floor by a string, applying a force of0.50 N. During the first meter, the string is para

llel to the floor. In the next two meters, the string makes an angle 0f 300 with the horizontal direction. What’s the total amount of work done by Richardson on the toy?
Physics
1 answer:
Anastasy [175]3 years ago
7 0

Answer:

Total Work done =0.65 joule

Explanation:

Work done is given Mathematically as

W=F *d

Where w=work done in joules

F=applied force

d= distance moved

The work done to move the toy accros the first meter is

W1=0.5*1

W1=0.5joule

The work done to move the toy across the next 2m at an angle of 30° is

.W2=0.5*2cos30

W2=0.5*2*0.154

W2=0.154joule

Hence total work done is

W1+W2=0.5+0.154

Total Work done =0.65 joule

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The first antiparticle, the positron or antielectron, was discovered in 1932. It had been predicted by Paul Dirac in 1928, thoug
Taya2010 [7]

Answer:

Energy of Photon = 4.091 MeV

Explanation:

From the conservation of energy principle, we know that total energy of the system must remain conserved. So, the energy or particles before collision must be equal to the energy of photons after collision.

K.E OF electron + Rest Energy of electron + K.E of positron + Rest Energy of positron = 2(Energy of Photon)

where,

K.E OF electron = 3.58 MeV

Rest Energy of electron = 0.511 MeV

Rest Energy of positron = 0.511 MeV

K.E OF positron = 3.58 MeV

Energy of Photon = ?

Therefore,

3.58 MeV + 0.511 MeV + 3.58 MeV + 0.511 MeV = 2(Energy of Photon)

Energy of Photon = 8.182 MeV/2

<u>Energy of Photon = 4.091 MeV</u>

8 0
3 years ago
What is formula of baking soda.​
Anika [276]

NaHCO3

That is the formula for baking soda

6 0
3 years ago
Read 2 more answers
An object whose weight is 10kg is placed on smooth plane inclined at 30° to the horizontal. find the acceleration of the object
velikii [3]

Answer:

4.9 m/s²

Explanation:

Draw a free body diagram.  There are two forces on the object:

Weight force mg pulling straight down,

and normal force N pushing perpendicular to the plane.

Sum the forces in the parallel direction.

∑F = ma

mg sin θ = ma

a = g sin θ

a = (9.8 m/s²) (sin 30°)

a = 4.9 m/s²

8 0
3 years ago
Two thin 80.0-cm rods are oriented at right angles to each other. Each rod has one end at the origin of the coordinates, and one
kogti [31]

Answer:

The net force on the electron is given as:

F = 1.35 x 10⁻¹³ N j - 1.35 x 10⁻¹³ N i

Explanation:

Given:

charge on rod along x-axis = Q₁ = -15 x 10⁻⁶ C

charge on rod along y-axis = Q₂ = 15 x 10⁻⁶ C

distance of electron from rod 1 = r₁ = 0.4 m

distance of electron from rod 1 = r₂ = 0.4 m

charge on electron = q = -1.6 x 10⁻¹⁹ C

ε° = 8.85 x 10⁻¹² C²/Nm²

Electric force on charge due to rod 1:

F₁ = qE = 1/4πε°(qQ₁/r₁²)

F₁ = (9 x 10⁹ x -1.6 x 10⁻¹⁹ x -15 x 10⁻⁶)/0.4²

F₁ = 1.35 x 10⁻¹³ N

Negative negative repels each other so the rod will Force the electron in positive y-direction.

F₁ = 1.35 x 10⁻¹³ N j

Electric force on charge due to rod 2:

F₂ = qE = 1/4πε°(qQ₂/r₂²)

F₂ = (9 x 10⁹ x -1.6 x 10⁻¹⁹ x 15 x 10⁻⁶)/0.4²

F₂ = - 1.35 x 10⁻¹³ N

Opposite charges attract each other so the rod will force the electron in negative x-direction.

F₂ =  - 1.35 x 10⁻¹³ N i

Net Force:

F = F₁ + F₂

F = 1.35 x 10⁻¹³ N j - 1.35 x 10⁻¹³ N i

4 0
3 years ago
Jupiter's semimajor axis is 7.78×1011 m. The mass of the Sun is 1.99×1030 kg. (a) What is the period of Jupiter's orbit in secon
dedylja [7]

Explanation:

It is given that,

Semi major axis of the Jupiter, a=7.78\times 10^{11}\ m

Mass of the sun, M=1.99\times 10^{30}\ kg

(a) Let T is the period of Jupiter's orbit. It is given by :

T^2\propto a^3

T^2=\dfrac{4\pi^2}{GM}a^3

T^2=\dfrac{4\pi^2}{6.67\times 10^{-11}\times 1.99\times 10^{30}}\times (7.78\times 10^{11})^3

T=3.74\times 10^8\ s

(b) We know that,

1\ year=3.154\times 10^7\ s

or

1\ s=3.171\times 10^{-8}\ year

3.74\times 10^8\ s={3.171\times 10^{-8}}\times {3.74\times 10^8}

T = 11.859 earth years

Hence, this is the required solution.

7 0
4 years ago
Read 2 more answers
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