Solution 1
Answer:
0.56
Explanation:
Let the coefficient of friction is μ.
m = 4.3 kg, θ = 30 degree, initial velocity, u = 0, s = 2.7 m, t = 5.8 s
By the free body diagram,
Normal reaction, N = mg Cosθ = 4.3 x 9.8 x Cos 30 = 36.49 N
Friction force, f = μ N = 36.49 μ
Net force acting on the block,
Fnet = mg Sinθ - f = 4.3 x 9.8 x Sin 30 - 36.49 μ
Fnet = 21.07 - 36.49μ
Net acceleartion, a = Fnet / m
a = (21.07 - 36.49μ) / 4.3
use second equation of motion
s = ut + 1/2 a t^2
2.7 = 0 + 1/2 x (21.07 - 36.49μ) x 5.8 x 5.8 / 4.3
By solving we get
μ = 0.56
📚 Related Questions
The work by tension is ¹¹/₁₀ Mgℓ
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Further explanation
Complete Question:
An object with mass M is attached to the end of a string and is raised vertically at a constant acceleration of g/10 . If it has been raised a distance ℓ from rest, how much work has been done by the tension in the string?
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Given:
Mass of the object = M
Acceleration of the object = g/10
Distance = ℓ
Asked:
Work by Tension = W = ?
Solution:
Let's find the magnitude of tension as follows:
[tex]\Sigma F = ma[/tex]
[tex]T - Mg = Ma[/tex]
[tex]T = Mg + Ma[/tex]
[tex]T = M(g + a)[/tex]
[tex]T = M(g + \frac{1}{10}g)[/tex]
[tex]T = M(\frac{11}{10}g)[/tex]
[tex]T = \frac{11}{10}Mg[/tex]
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[tex]W = T \times L[/tex]
[tex]W = \frac{11}{10}Mg \times L[/tex]
[tex]W = \frac{11}{10}MgL[/tex]
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The work by tension is ¹¹/₁₀ Mgℓ
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Answer details
Grade: High School
Subject: Physics
Chapter: Dynamics
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Keywords: Gravity , Unit , Magnitude , Attraction , Distance , Mass , Newton , Law , Gravitational , Constant
Answer:
The work done by tension in string is [tex]\dfrac{11}{10}MgL}[/tex].
Explanation:
Given data:
Mass of object is, M.
Acceleration of object is, a = g/10.
Distance covered vertically is, L.
The work done by tension in the string is given as,
[tex]W = T \times L[/tex] .......................................................... (1)
Here, T is the tension force on string.
Apply the equilibrium of forces on string as,
[tex]T- Mg=Ma[/tex]
Here, g is gravitational acceleration.
[tex]T- Mg=M(\dfrac{g}{10} )\\T=M(\dfrac{g}{10} )+Mg\\T=\dfrac{11}{10}Mg[/tex]
Substituting value in equation (1) as,
[tex]W = \dfrac{11}{10}Mg \times L\\W = \dfrac{11}{10}MgL}[/tex]
Thus, the work done by tension in string is [tex]\dfrac{11}{10}MgL}[/tex].
For more details, refer to the link:
https://brainly.com/question/14871010?referrer=searchResults
Answer:
Angle of ray makes with the vertical is 62.1 degree
Explanation:
As per the ray diagram we know that the angle of incidence on oil brine interface will be given as
[tex]tan\theta_i = \frac{0.7}{0.98}[/tex]
[tex]\theta_i = 35.5^0[/tex]
now by Snell'a law at that interface we have
[tex]\mu_1 sin\theta_i = \mu_2 sin \theta_r[/tex]
now we will have
[tex]1.52 sin35.5 = 1.40 sin\theta_r[/tex]
[tex]\theta_r = 39.12^0[/tex]
now this is the angle of incidence for oil air interface
so now again by Snell's law we will have
[tex]\mu_2 sin\theta_i' = \mu_{air} sin\theta[/tex]
[tex]1.40 sin39.12 = 1 sin\theta[/tex]
[tex]\theta = 62.1^0[/tex]
Answer:
Tension, T = 1736 N
Explanation:
It is given that,
Mass of bricks, m = 175 kg
A rope is attached to a load of 175 kg bricks lifts the bricks with a steady acceleration of 0.12 m/s² in vertically upwards direction. let T is the tension in the rope. Using second equation of motion as :
T - mg = ma
T = ma + mg
T = m(a + g)
T = 175 kg ( 0.12 m/s² + 9.8 m/s² )
T = 1736 N
Hence, the tension in the wire is 1736 N.
Answer:
The tension in the rope is 1736 N.
(e) is correct option.
Explanation:
Given that,
Mass of bricks = 175 kg
Acceleration = 0.12 m/s²
Let T is the tension in the rope.
A rope attached to a load of 175 kg bricks lifts the bricks with a steady acceleration of 0.12.m/s^2 in vertically upward direction.
Using equation of balance
[tex]T-mg=ma[/tex]
[tex]T=mg+ma[/tex]
[tex]T=175(9.8+0.12)[/tex]
[tex]T= 1736\ N[/tex]
Hence, The tension in the rope is 1736 N.
Answer:
see attachment
Explanation:
The electric potential varies inversely with the distance. So, the third charge should be placed at a distance of 3 cm from the origin on the x-axis.
What is electric potential?
The work done on an electric charge to shift it from infinity to a point is known as electric potential at that point. And its expression is,
[tex]V = \dfrac{kq}{r}[/tex]
here, k is the coulomb's constant.
Given data:
The magnitude of two point charges are, [tex]+2.0 \;\rm \mu C[/tex] and [tex]-6.0 \;\rm \mu C[/tex].
The location of each charge on the x-axis is -1.0 cm and +2.0 cm.
Let the third charge ( [tex]+3.0 \;\rm \mu C[/tex] ) be placed at a distance of x. Then the electric potential at origin is,
[tex]V = \dfrac{k \times 2.0}{0.01} +\dfrac{k \times -6.0}{0.02} +\dfrac{k \times 3.0}{x}[/tex]
Since, potential at origin is zero (V = 0). Then,
[tex]0 = \dfrac{k \times 2.0}{0.01} +\dfrac{k \times -6.0}{0.02} +\dfrac{k \times 3.0}{x}\\\\\dfrac{k \times 6.0}{0.02} = \dfrac{k \times 2.0}{0.01} +\dfrac{k \times 3.0}{x}\\\\\dfrac{6.0}{0.02} = \dfrac{2.0}{0.01} +\dfrac{3.0}{x}\\\\x = 0.03 \;\rm m =3 \;\rm cm[/tex]
Thus, we can conclude that the third charge should be placed at a distance of 3 cm from the origin on the x-axis.
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you use the formula F=ma to get the answer
Answer:
the answer is in the picture below......
(a) Calculate the ratio of the electric potential at the surface of the large sphere to that of the small sphere.
Let the bigger sphere be sphere 1 and the let the smaller sphere be sphere 2. Rest of the answer is in the picture.
Answer:
V=-5304.6V
Explanation:
(I set the charge density unit to microcoulombs per metre cubed)
Answer:
In the picture.
(a) What is the impulse delivered to the ball by the racket?
(b) What work does the racket do on the ball? (Indicate the direction with the sign of your answer.)
(a) The impulse delivered to the ball by the racket is 5.24 kg.m/s
(b) The work that the racket does on the ball is -35.1 Joule
Further Explanation
Given :
mass of ball = m = 0.06 kg
initial velocity = v₁ = -50.4 m/s
final velocity = v₂ = 37.0 m/s
Unknown :
(a) Impulse = I = ?
(b) Work = W = ?
Solution :
Question (a) :
In this question , we could use the formula from Second Law of Newton :
[tex]I = \Delta p[/tex]
[tex]I = p_2 - p_1[/tex]
[tex]I = m \times v_2 - m \times v_1[/tex]
[tex]I = m \times (v_2 - v_1)[/tex]
[tex]I = 0.06 \times (37.0 - (-50.4))[/tex]
[tex]I = 0.06 \times (87.4)[/tex]
[tex]I = 5.244~kg.m/s[/tex]
[tex]\large { \boxed {I \approx 5.24~kg.m/s} }[/tex]
Question (b) :
[tex]W = F \times d[/tex]
[tex]W = (\frac{I}{\Delta t})(\frac {v_1 + v_2}{2} \Delta t)[/tex]
[tex]W = \frac{I(v_1 + v_2)}{2}[/tex]
[tex]W = \frac{5.244(-50.4 + 37)}{2}[/tex]
[tex]W = \frac{5.244(-13.4)}{2}[/tex]
[tex]W = -35.1348~Joule[/tex]
[tex]\large { \boxed {W \approx -35.1~Joule}}[/tex]
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Answer details
Grade: High School
Subject: Physics
Chapter: Dynamics
Keywords: Newton, Law, Impulse, Work
Explanation and answer:
This problem is best answered by drawing a figure as a first step.
ABC is the scaffold.
A downward force of 500N is applied downwards at 1m from end A.
The weight of 800N is exerted by the scaffold uniformly distributed between A & C.
At A and C, ropes are attached to support the load.
Let Fc=tension in rope passing through C.
Take moments about A:
Fc = (500N * 1m +800N*(3+1)/2m / 4m
= (500 Nm + 1600Nm) / 4m
= 2100 Nm / 4m
= 525 N