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ollegr [7]
3 years ago
8

A child pulls a wagon at a constant velocity along a level sidewalk. The child does this by applying a 22 newton force to the wa

gon handle, which is inclined at 35° to the sidewalk as shown below. What is the magnitude of the fore of friction on the wagon?
Physics
1 answer:
Jlenok [28]3 years ago
5 0

Answer:

The answer is 18 N.

Explanation:

A force can be divided into components x and y components. The component along the x-axis is called the horizontal component and along the y-axis is called the vertical component. In this case, as the force is in a horizontal direction and is also known as x-component of force. The x- component of force is  

Fx = Fcosθ

Fx = 22(cos 35°)

Fx = 22 x 0.819

Fx =  18 N

Child's horizontal pull forces are equal to that of frictional resistance force on the wagon.

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What is the value of Vo in the
notsponge [240]

Answer: 60m/s

Explanation:

From the diagram:

Θ = 30°

Vertical resolution (y-axis) :

Voy = VoSinΘ

g in the upward direction = negative (-) = - g

Vfinal = 0

Distance (H) traveled along y =

Time taken to reach maximum height :

From v = u + at

0 = usinΘ - gt

gt = usinΘ

t = usinΘ / g

Horizontal resolution:

S = ut + 1/2at^2

Substituting t = usinΘ / g ; Voy = usinΘ

S = (usinΘ × usinΘ / g) - 1/2 g × (usinΘ /g)^2

S = (u^2sin^2Θ / g) - (u^2sin^2Θ / 2g)

S = (u^2sin^2Θ) / 2g

Now if S = maximum height = 45m

Then,

45 = [Vo^2sin^2(30°)] / 2(10)

45 =[ Vo^2 * (0.5)^2] / 20

45 =( Vo^2 * 0.25) / 20

20 * 45 = Vo^2 * 0.25

900 / 0.25 = Vo^2

3600 = Vo^2

Vo = sqrt(3600)

Vo = 60m/s

5 0
3 years ago
A particle that has mass m and charge q enters a uniform magnetic field that has magnitude B and is directed along the x axis. T
Usimov [2.4K]

Answer:

a) The trajectory will be a helical path.

b) θ = 2*π rad

Explanation:

a) Since the initial velocity of the particle has a component parallel (x-component) to the magnetic  field B , the trajectory will be a helical path.

b) Given

t = 2*π*m/(q*B)

We can use the equation

θ = ω*Δt

where

θ is the angular displacement

ω is the angular speed, which is obtained as follows:

ω = q*B/m

then we have

θ = (q*B/m)*2*π*m/(q*B)

⇒  θ = 2*π rad

6 0
3 years ago
A scientist is subjected to a dose of ionizing radiation in his laboratory. without the help of radiation detection instruments,
vesna_86 [32]
<span>Health problems that develop later</span>
5 0
3 years ago
Two 2.0 kg bodies, A and B, collide. The velocities before the collision are ~vA = (15ˆi + 30ˆj) m/s and ~vB = (−10ˆi + 5.0ˆj) m
AleksandrR [38]

Answer:

Part a)

10\hat i + 15\hat j = \vec v

Part b)

\Delta K = 500 J

Explanation:

As we know that there is no external force on the system of two masses so here total momentum of the system will remains conserved

so we can say

m_1v_{1i} + m_2v_{2i} = m_1v_{1f} + m_2v_{2f}

(2kg)(15\hat i + 30 \hat j) + (2 kg)(-10\hat i  + 5\hat j) = 2kg(-5\hat i + 20\hat j) + 2\vec v

5\hat i + 35\hat j = (-5\hat i + 20\hat j) +\vec v

10\hat i + 15\hat j = \vec v

Part b)

magnitude of the initial speed of A = \sqrt{15^2 + 30^2} = 33.54 m/s

magnitude of the initial speed of B = \sqrt{10^2 + 5^2} = 11.18 m/s

magnitude of final speed of A = \sqrt{5^2 + 20^2} = 20.61 m/s

magnitude of final speed of B = \sqrt{10^2 + 15^2} = 18.03 m/s

Now change in total kinetic energy is given as

\Delta K = (\frac{1}{2}m_1v_{1i}^2 + \frac{1}{2}m_2v_{2i}^2) - (\frac{1}{2}m_1v_{1f}^2 + \frac{1}{2}m_2v_{2f}^2)

\Delta K = (\frac{1}{2}2(33.54)^2 + \frac{1}{2}2(11.18)^2) - (\frac{1}{2}2(20.61)^2 + \frac{1}{2}2(18.03)^2)

\Delta K = 500 J

6 0
3 years ago
(a) Calculate the self-inductance (in mH) of a 55.0 cm long, 10.0 cm diameter solenoid having 1000 loops.
DedPeter [7]

Explanation:

(a) We have,

Length of solenoid, l = 55 cm = 0.55 m

Diameter of the solenoid, d = 10 cm

Radius, r = 5 cm = 0.05 m

Number of loops in the solenoid is 1000.

(a) The self inductance in the solenoid is given by :

L=\dfrac{\mu_o N^2A}{l}

A is area

L=\dfrac{4\pi \times 10^{-7}\times (1000)^2\times \pi (0.05)^2}{0.55}\\\\L=0.0179\ H\\\\L=17.9\ mH

(b) The energy stored in the inductor is given by :

E=\dfrac{1}{2}LI^2\\\\E=\dfrac{1}{2}\times 0.0179\times (19.5)^2\\\\E=3.4\ J

Hence, this is the required solution.

4 0
3 years ago
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