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Eduardwww [97]
2 years ago
15

13–82. the 8-kg sack slides down the smooth ramp. if it has a speed of 1.5 m> s when y = 0.2 m, determine the normal reaction

the ramp exerts on the sack and the rate of increase in the speed of sack at this instant.
y
*13–84. the 2-lb block is released from rest at a and slides down along the smooth cylindrical surface. if the attached spring has a stiffness k = 2 lb> ft, determine its unstretched length so that it does not allow the block to leave the surface until u = 60°.
a
prob. 13–82
prob. 13–84
2,fuuu 2,frur
Physics
1 answer:
marissa [1.9K]2 years ago
4 0
The second problem requires a figure to be answered. For the first problem
The acceleration of the sack is
1.5² - 0² = 2a(0.2)
a = 5.63 m/s2
The reaction of the ramp is
F = 8 kg (5.63 m/s2)
F = 45 N
Differentiate the kinematic equation involving time to get the rate of increase of the velocity.
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Michael Phelps needs to swim at an average speed of 2.00 m/s in order to set a new world record in the 200 m freestyle. If he sw
Natasha_Volkova [10]

Answer:

Explanation:

Given

average speed of Phelps v_{avg}=2\ m/s

total distance d=200\ m

he swims first 100 m at an average speed if 1.8 m/s

so time taken is t_1=\frac{100}{1.8}=55.55\ s

suppose t_2 is the time taken to swim remaining half

average velocity is v_{avg}=\frac{displacement}{total\ time}

v_{avg}=\frac{100+100}{55.55+t_2}

t_2+55.55=\frac{200}{2}=100

t_2=44.45\ s

so velocity in the second half is

v_2=\frac{100}{45.45}

v_2=2.19\approx 2.2\ m/s                                      

3 0
1 year ago
would an elephant standing on one leg exert a higher force on a scale than an elephant on four legs. why​
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Answer:

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7 0
2 years ago
What is the torque τb about axis b due to the force f⃗ ? (b is the point at cartesian coordinates (0,b), located a distance b fr
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Check the attached file for the solution.

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Describe a theme-park ride that has constant speed but changing velocity.<br><br> Thank you!
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8 0
2 years ago
A charge of 8.4 × 10–4 C moves at an angle of 35° to a magnetic field that has a field strength of 6.7 × 10–3 T. If the magnetic
larisa86 [58]

Answer:

The charge is moving with the  velocity of 1.1\times10^{4}\ m/s.

Explanation:

Given that,

Charge q =8.4\times10^{-4}\ C

Angle = 35°

Magnetic field strength B=6.7\times10^{-3}\ T

Magnetic force F=3.5\times10^{-2}\ N

We need to calculate the velocity.

The Lorentz force exerted by the magnetic field on a moving charge.

The magnetic force is defined as:

F = qvB\sin\theta

v = \dfrac{F}{qB\sin\theta}

Where,

F =  Magnetic force

q = charge

B = Magnetic field strength

v = velocity

Put the value into the formula

v =\dfrac{3.5\times10^{-2}}{8.4\times10^{-4}\times6.7\times10^{-3}\times\sin35^{\circ}}

v =\dfrac{3.5\times10^{-2}}{8.4\times10^{-4}\times6.7\times10^{-3}\times0.57}

v = 10910.36\ m/s

v = 1.1\times10^{4}\ m/s

Hence, The charge is moving with the  velocity of 1.1\times10^{4}\ m/s.

4 0
1 year ago
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