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omeli [17]
3 years ago
9

A 5.5kg mass is pushed with a force of 31N across a table having μk of 0.350. Find how fast it will accelerate, taking friction

into consideration.

Physics
1 answer:
Ipatiy [6.2K]3 years ago
4 0

Answer:

3.66m/s^2

Explanation:

First, we need to find the friction which is F=u*N.

After that, find the resultant force by substarcting the friction from the forward force.

Lastly,using the formula F=ma, substitute in the known values of F and m to find a

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You are watching an archery tournament when you start wondering how fast an arrow is shot from the bow. Remembering your physics
spayn [35]

Answer:

v_0 = 3.53~{\rm m/s}

Explanation:

This is a projectile motion problem. We will first separate the motion into x- and y-components, apply the equations of kinematics separately, then we will combine them to find the initial velocity.

The initial velocity is in the x-direction, and there is no acceleration in the x-direction.

On the other hand, there no initial velocity in the y-component, so the arrow is basically in free-fall.

Applying the equations of kinematics in the x-direction gives

x - x_0 = v_{x_0} t + \frac{1}{2}a_x t^2\\63 \times 10^{-3} = v_0t + 0\\t = \frac{63\times 10^{-3}}{v_0}

For the y-direction gives

v_y = v_{y_0} + a_y t\\v_y = 0 -9.8t\\v_y = -9.8t

Combining both equation yields the y_component of the final velocity

v_y = -9.8(\frac{63\times 10^{-3}}{v_0}) = -\frac{0.61}{v_0}

Since we know the angle between the x- and y-components of the final velocity, which is 180° - 2.8° = 177.2°, we can calculate the initial velocity.

\tan(\theta) = \frac{v_y}{v_x}\\\tan(177.2^\circ) = -0.0489 = \frac{v_y}{v_0} = \frac{-0.61/v_0}{v_0} = -\frac{0.61}{v_0^2}\\v_0 = 3.53~{\rm m/s}

6 0
3 years ago
A plate in a parallel-plate capacitor has an area of 0.03 m2 and is 0.5 × 10–3 m from the other plate. The space between the pla
aivan3 [116]

Answer:

4 x 10^-9F

Explanation:

7 0
3 years ago
So this helicopter pilot dropped me in the middle of an absolutely smooth frictionless
Llana [10]
Dang dude you are a soldier! Good job
5 0
2 years ago
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Two ropes have equal length and are stretched the same way. The speed of a pulse on rope 1 is 1.4 times the speed on rope 2. Par
kondor19780726 [428]

Answer:

m1/m2 = 0.51

Explanation:

First to all, let's gather the data. We know that both rods, have the same length. Now, the expression to use here is the following:

V = √F/u

This is the equation that describes the relation between speed of a pulse and a force exerted on it.

the value of "u" is:

u = m/L

Where m is the mass of the rod, and L the length.

Now, for the rod 1:

V1 = √F/u1 (1)

rod 2:

V2 = √F/u2 (2)

Now, let's express V1 in function of V2, because we know that V1 is 1.4 times the speed of rod 2, so, V1 = 1.4V2. Replacing in the equation (1) we have:

1.4V2 = √F/u1 (3)

Replacing (2) in (3):

1.4(√F/u2) = √F/u1 (4)

Now, let's solve the equation 4:

[1.4(√F/u2)]² = F/u1

1.96(F/u2) =F/u1

1.96F = F*u2/u1

1.96 = u2/u1 (5)

Now, replacing the expression of u into (5) we have the following:

1.96 = m2/L / m1/L

1.96 = m2/m1 (6)

But we need m1/m2 so:

1.96m1 = m2

m1/m2 = 1/1.96

m1/m2 = 0.51

5 0
3 years ago
What can you say about the magnitudes of the forces that the balloons exert on each other?
maxonik [38]

Answer:

F_G=G. \frac{m_1.m_2}{R^2} gravitational force

F=\frac{1}{4\pi \epsilon_0} \times \frac{q_1.q_2}{R^2} electrostatic force

Explanation:

The forces that balloons may exert on each other can be gravitational pull due to the mass of the balloon membrane and the mass of the gas contained in each. This force is inversely proportional to the square of the radial distance between their center of masses.

The Mutual force of gravitational pull that they exert on each other can be given as:

F_G=G. \frac{m_1.m_2}{R^2}

where:

G= gravitational constant  =6.67\times 10^{-11} m^3.kg^{-1}.s^{-2}

m_1\ \&\ m_2 are the masses of individual balloons

R= the radial distance between the  center of masses of the balloons.

But when  there are charges on the balloons, the electrostatic force comes into act which is governed by Coulomb's law.

Given as:

F=\frac{1}{4\pi \epsilon_0} \times \frac{q_1.q_2}{R^2}

where:

\rm \epsilon_0= permittivity\ of\ free\ space

q_1\ \&\ q_2 are the charges on the individual balloons

R = radial distance between the charges.

3 0
3 years ago
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