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Degger [83]
2 years ago
9

A child is sliding a toy block (with mass = m) down a ramp. The coefficient of static friction between the block and the ramp is

0.25. When the block is halfway down the ramp, the child pushes down on the block perpendicular to the plane, halting it. What is the minimum force the child must apply to keep the block from starting to slide down the ramp?

Physics
2 answers:
tiny-mole [99]2 years ago
8 0

Answer:

F=mg(sin(\theta )-0.25 cos(\theta ))

Explanation:

The free body diagram of the block on the slide is shown in the below figure

Since the block is in equilibrium we apply equations of statics to compute the necessary unknown forces

N is the reaction force between the block and the slide

For equilibrium along x-axis we have

\sum F_{x}=0\\\\mgsin(\theta )-\mu N-F=0\\\therefore F=mgsin(\theta)-\mu N......(\alpha )\\Similarly\\\sum F_{y}=0\\\\N-mgcos(\theta )=0\\\therefore N=mgcos(\theta ).......(\beta )\\\\

Using value of N from equation β in α we get value of force as

F=mg(sin(\theta )-\mu cos(\theta ))

Applying values we get

F=mg(sin(\theta )-0.25 cos(\theta ))

nika2105 [10]2 years ago
3 0

Answer:

F = \frac{mgsin\theta}{\mu} - mgcos\theta

Explanation:

As we know that the force applied is perpendicular to inclined plane

So here the normal force on the block is given as

N = F + mgcos\theta

now in order to stop the block by the perpendicular force we can say that the friction force is sufficiently large to balance the force of gravity

So here we can say

F_f = mgsin\theta

\mu N = mgsin\theta

\mu(F + mg cos\theta) = mg sin\theta

now we have

F = \frac{mgsin\theta}{\mu} - mgcos\theta

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omeli [17]

Answer:

0.0016 cm

Explanation:

\alpha_b = Thermal coefficient of expansion of brass = 19\times 10^{-6}\ /^{\circ}C

\alpha_g = Thermal coefficient of expansion of glass = 9\times 10^{-6}\ /^{\circ}C

\Delta T = Change in temperature = (60-20)^{\circ}C

R_0 = Initial radius = 4 cm

Change in radius of material is given by

R=R_0(1+\alpha\Delta T)

Difference in radii of the lid and jar

\Delta R=R_b-R_g\\\Rightarrow \Delta R=R_0(1+\alpha_b\Delta T)-R_0(1+\alpha_g\Delta T)\\\Rightarrow \Delta R=R_0(\alpha_b-\alpha_g)\Delta T\\\Rightarrow \Delta R=4\times (19\times 10^{-6}-9\times 10^{-6})\times (60-20)\\\Rightarrow \Delta R=0.0016\ cm

The size of the gap is 0.0016 cm or 0.000016 m

8 0
2 years ago
pitot tube on an airplane flying at a standard sea level reads 1.07 x 105 N/m2. What is the velocity of the airplane?
Allushta [10]

Answer:

V_infinty=98.772 m/s

Explanation:

complete question is:

The following problem assume an inviscid, incompressible flow. Also, standard sea level density and pressure are 1.23kg/m3(0.002377slug/ft3) and 1.01imes105N/m2(2116lb/ft2), respectively. A Pitot tube on an airplane flying at standard sea level reads 1.07imes105N/m2. What is the velocity of the airplane?

<u>solution:</u>

<u>given:</u>

<em>p_o=1.07*10^5 N/m^2</em>

<em>ρ_infinity=1.23 kg/m^2</em>

<em>p_infinity=1.01*10^5 N/m^2</em>

p_o=p_infinity+(1/2)*(ρ_infinity)*V_infinty^2

V_infinty^2=9756.097

V_infinty=98.772 m/s

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zepelin [54]

Answer:

q2 must also be doubled

r may also be halved

Explanation:

According to Coulumbs law

F= K q1 q2/r^2

If q1 is doubled, we must necessarily double q2 and r may also be halved in order to maintain F at the same value. Once the value of F is thus kept constant and E is also constant, the product FE must remain constant.

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alekssr [168]
D
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avanturin [10]
Weight expressed in Newtons is expressed in the equation whereby Weight= the mass of an object * the force of gravity. The force of gravity on earth is a constant 9.8 meters per second squared. Therefore if weight (w) = 63 N and the force of gravity is 63 N then the mass must equal 6.43 kg. Because the equation for weight is w=mg so 63 N (w) = m * 9.8 m/s^2. 
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