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zhannawk [14.2K]
1 year ago
13

Consider another special case in which the inclined plane is vertical (θ=π/2). In this case, for what value of m1 would the acce

leration of the two blocks be equal to zero? Express your answer in terms of some or all of the variables m2 and g.

Physics
1 answer:
Lana71 [14]1 year ago
6 0

Answer:

Explanation:

Consider another special case in which the inclined plane is vertical (θ=π/2). In this case, for what value of m1 would the acceleration of the two blocks be equal to zero

F - Force

T = Tension

m = mass

a = acceleration

g = gravitational force

Let the  given Normal on block 2 = N

and N = m_2 g \cos \theta

and the tension in the given string is said to be T = m_2 g \sin \theta

When the acceleration a=\frac{F}{m_1}

for the said block 1.

It will definite be zero only when Force is zero , F=0.

Here by Force, F

I refer net force on block 1.

Now we know

F = m_1g-T.

It is known that if the said

\theta=\frac{\pi}{2} ,

then Tension T= m_2g [since \sin(\pi/2) = 1],

Now making "F = m_1g - m_2g"

So If we are to make Force equal to zero

F=0 => m_1g = m_2g \ or \ m_1 = m_2

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A steel cable 1.25 in. in diameter and 50 ft long is to lift a 20-ton load without permanently deforming. What is the length of
Over [174]

Answer:

50.0543248872 ft

Explanation:

F = Load = 20 ton = 20\times 2000\ lb

d = Diameter = 1.25 in

L_1 = Initial length = 50 ft

L_2 = Final length

A = Area = \dfrac{\pi}{4}d^2

Y = Young's modulus = 30\times 10^6\ psi

Young's modulus is given by

Y=\dfrac{FL}{A\Delta L}\\\Rightarrow Y=\dfrac{FL_1}{\dfrac{\pi}{4}d^2(L_2-L_1)}\\\Rightarrow L_2=\dfrac{4FL_1}{Y\pi d^2}+L_1\\\Rightarrow L_2=\dfrac{4\times 40000\times 50}{30\times 10^6\times \pi\times 1.25^2}+50\\\Rightarrow L_2=50.0543248872\ ft

The length during the lift is 50.0543248872 ft

6 0
1 year ago
An electron is located in an electric field of magnitude 600. newtons per coulomb. what is the magnitude of the electrostatic fo
SOVA2 [1]
The magnitude of the electrostatic force acting on a charge q is the product between the charge and the intensity of the electric field E. The magnitude of the electron charge is q=1.6 \cdot 10^{-19}C (we are not interested in the sign), so the electrostatic force magnitude is
F=qE=(1.6 \cdot 10^{-19}C)(600 N/C)=9.6\cdot 10^{-17}N
6 0
1 year ago
A frog leaps up from the ground and lands on a step 0.1 m above the ground 2 s later. We want to find the vertical velocity of t
zhuklara [117]

Answer:

<em>You would use the kinematic formula:</em>

    \Delta y=V_{0y}\times t-g\times t^2/2

Explanation:

The upwards vertical motion is ruled by the equation:

        y=y_0+V_{0y}\times t-g\times t^2/2

Where:

       y \text{ is the position at the time }t:y=0.1m

       y_0\text{ is the initial position: }y_0=0

       t=2s

       g\text{ is the gravitational acceleration: }\approx 9.8m/s^2

       V_{0y}\text{ is the initial vertical velocity}

Naming Δy = y - y₀, the equation becomes:

      \Delta y=V_{0y}\times t-g\times t^2/2

Then, you just need to substitute with Δy = 0.1m, t = 2s, and g = 9.8m/s², ans solve for the intital vertical velocity.

7 0
1 year ago
Read 2 more answers
. A girl runs and jumps horizontally off a platform 10m above a pool with a speed of 4.0m/s. As soon as she leaves the platform,
faust18 [17]

Answer:

2.39 revolutions

Explanation:

As she jumps off the platform horizontally at a speed of 10m/s, the gravity is the only thing that affects her motion vertically. Let g = 10m/s2, the time it takes for her to fall 10m to water is

h = gt^2/2

10 = 10t^2/2

t^2 = 2

t = \sqrt{2} = 1.414 s

Knowing the time it takes to fall to the pool, we calculate the angular distance that she would make at a constant acceleration of 15 rad/s2:

\theta = \alpha t^2/2

\theta = 15 * 2/2 = 15 rad

As each revolution is 2π, the total number of revolution that she could make is: 15 / 2π = 2.39 rev

3 0
2 years ago
What is the total kinetic energy of a 0.15 kg hockey puck sliding at 0.5 m/s and rotating about its center at 8.4 rad/s? The dia
ycow [4]
The mass of the puck is
m = 0.15 kg.
The diameter of the puck is 0.076 m, therefore its radius  is
r = 0.076/2 = 0.038 m
The sliding speed is
v = 0.5 m/s
The angular velocity is
ω = 8.4 rad/s

The rotational moment of inertia of the puck is
I = (mr²)/2
  = 0.5*(0.15 kg)*(0.038 m)²
  = 1.083 x 10⁻⁴ kg-m²

The kinetic energy of the puck is the sum of the translational and rotational kinetic energy.
The translational KE is
KE₁ = (1/2)*m*v²
       = 0.5*(0.15 kg)*(0.5 m/s)²
       = 0.0187 j

The rotational KE is
KE₂ = (1/2)*I*ω²
       = 0.5*(1.083 x 10⁻⁴ kg-m²)*(8.4 rad/s)²
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The total KE is
KE = 0.0187 + 0.0038 = 0.0226 J

Answer: 0.0226 J


4 0
1 year ago
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