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Xelga [282]
3 years ago
13

The box leaves position x=0 with speed v0. The box is slowed by a constant frictional force until it comes to rest at position x

=x1. Find Ff, the magnitude of the average frictional force that acts on the box. (Since you don't know the coefficient of friction, don't include it in your answer.)
Physics
1 answer:
Diano4ka-milaya [45]3 years ago
4 0

Answer:

Ff=m\times \dfrac{V_o^2}{2X_1}

Explanation:

Given that

At X=0 V=Vo

At X=X1  V=0

As we know that friction force is always try to oppose the motion of an object. It means that it provide acceleration in the negative direction.

We know that

V^2=U^2-2aS

0=V_o^2-2a X_1

a=\dfrac{V_o^2}{2X_1}

So the friction force on the box

Ff= m x a

Ff=m\times \dfrac{V_o^2}{2X_1}

Where m is the mass of the box.

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Magnetic resonance imaging (mri) depends on the presence in tissue of __________________.
storchak [24]

Answer:

Pretty sure magnetic waves thank me later bro bro

8 0
2 years ago
Two objects (42.0 and 21.0 kg) are connected by a massless string that passes over a massless, frictionless pulley. The pulley h
Kazeer [188]

Answer:

a = 3.27 m/s²

T = 275 N

Explanation:

Given that:

Mass m₁ = 42.p0 kg

Mass m₂ = 21.0 kg

Consider both masses to be in a whole system, then:

The acceleration can be determined as:

(m_1+m_2)a = g(m_1-m_2)

Making acceleration the subject in the above formula;

a =\dfrac{g(m_1-m_2)}{(m_1+m_2)}

a =\dfrac{9.8(42.0-21.0)}{(42.0+21.0)}

a =\dfrac{9.8(21.0)}{(63.0)}

a =\dfrac{205.8}{(63.0)}

a = 3.27 m/s²

in the string, the tension is calculated using the formula:

T = \dfrac{2m_1m_2g}{(m_1+m_2)}

T = \dfrac{2(42)(21)(9.81)}{(42+21)}

T = \dfrac{17304.84}{63}

T = 274.68 N

T ≅ 275 N

8 0
3 years ago
A straight wire of length 0.62 m carries a conventional current of 0.7 amperes. What is the magnitude of the magnetic field made
anyanavicka [17]

Answer:

Magnetic field at point having a distance of 2 cm from wire is 6.99 x 10⁻⁶ T

Explanation:

Magnetic field due to finite straight wire at a point perpendicular to the wire is given by the relation :

B=\frac{\mu_{0}I }{2\pi R }\times\frac{L}{\sqrt{L^{2}+R^{2}  } }      ......(1)

Here I is current in the wire, L is the length of the wire, R is the distance of the point from the wire and μ₀ is vacuum permeability constant.

In this problem,

Current, I = 0.7 A

Length of wire, L = 0.62 m

Distance of point from wire, R = 2 cm = 2 x 10⁻² m = 0.02 m

Vacuum permeability, μ₀ = 4π x 10⁻⁷ H/m

Substitute these values in equation (1).

B=\frac{4\pi\times10^{-7}\times  0.7 }{2\pi \times0.02 }\times\frac{0.62}{\sqrt{(0.62)^{2}+(0.02) ^{2}  } }

B = 6.99 x 10⁻⁶ T

3 0
3 years ago
A river flows due east at 1.70 m/s. A boat crosses the river from the south shore to the north shore by maintaining a constant v
ad-work [718]

Answer:

a)

v = 14.1028 m/s  

∅ = 83.0765° north of east

b)

the required distance is 40.98 m

Explanation:

Given that;

velocity of the river u = 1.70 m/s

velocity of boat v = 14.0 m/s

Now to get the velocity of the boat relative to shore;

( north of east), we say

a² + b² = c²

(1.70)² + (14.0)² = c²

2.89 + 196 = c²

198.89 = c²

c = √198.89

c = 14.1028 m/s  

tan∅ = v/u = 14 / 1.7 =  8.23529

∅ = tan⁻¹ ( 8.23529 ) = 83.0765° north of east

Therefore, the velocity of the boat relative to shore is;

v = 14.1028 m/s  

∅ = 83.0765° north of east

b)  

width of river = 340 m,

ow far downstream has the boat moved by the time it reaches the north shore in meters = ?

we say;

340sin( 90° - 83.0765°)

⇒ 340sin( 6.9235°)

= 40.98 m

Therefore, the required distance is 40.98 m

5 0
3 years ago
Sometimes in the winter, ice crystals grow directly from the moisture in the air. As the water changes from a gas to a solid, th
mrs_skeptik [129]

Answer: B. slow down as they lose energy

7 0
3 years ago
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