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11 days ago

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On the earth, when an astronaut throws a 0.250-kg stone vertically upward, it returns to his hand a time T later. On planet X he

finds that, under the same circumstances, the stone returns to his hand in 2T. In both cases, he throws the stone with the same initial velocity and it feels negligible air resistance. The acceleration due to gravity on planet X (in terms of g) is _______.a) g/. b) g/4. c) 2g. d) g/2. e) g.

1 answer:

inna [987]11 days ago

4 0

Answer:

The correct option is d) a ’= g / 2

Explanation:

To address this scenario, we will apply the equations of kinematics.

On Earth:

v = v₀ - a t

a = (v₀- v) / T

On Planet X:

v = v₀ - a' t’

a ’= (v₀-v) / 2T

Let’s insert the terrestrial values into the model for Planet X:

a’= a / 2

Now applying Newton's second law:

W = ma

m g = m a

a = g

Upon substitution:

a ’= g / 2

Thus, we conclude that on Planet X, the gravitational acceleration is half that of Earth's.

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A uniform 1.0-N meter stick is suspended horizontally by vertical strings attached at each end. A 2.0-N weight is suspended from

serg [1189]

Response:

3.5 N

Reasoning:

Taking the 0 cm position as the pivot point, to achieve balance, the total moment calculated around this point must equal zero. We will analyze the moment generated at each point, moving from 0 to 100 cm:

- Tension from the string at the 0 cm mark is 0, since the moment arm is nonexistent.

- A 2 N weight at the 10 cm point creates a moment of 20 Ncm moving clockwise.

- Another 2 N weight at the 50 cm position produces a moment of 100 Ncm also clockwise.

- The weight of the 1 N stick located at its center of mass (50 cm) has a moment of 50 Ncm, clockwise.

- A 3 N weight at the 60 cm position generates a moment of 180 Ncm, clockwise.

- The tension T (N) from the string at the 100 cm end contributes a moment of 100T Ncm, moving counter-clockwise.

Total clockwise moments = 20 + 100 + 50 + 180 = 350 Ncm.

Total counter-clockwise moment = 100T.

To achieve balance, we set 100 T = 350, leading to T = 350 / 100 = 3.5 N.

4 0

6 days ago

Three point charges are positioned on the x axis. If the charges and corresponding positions are +32 µC at x = 0, +20 µC at x =

Sav [1095]

Response:

Magnitude of the electrostatic force acting on the +32 µC charge, $F_{net} = 12 N$

Clarification:

Let q₁ = +32 µC, located at x₁ = 0

q₂ = +20 µC, positioned at x₂ = 40 cm = 0.4 m

q₃ = -60 µC, placed at x₃ = 60 cm = 0.6 m

Define the force magnitude on the +32 µC charge from the +20 µC charge as F₁ (the force on q₁ due to q₂).

$F_{2} = \frac{kq_{1}q_{2} }{x_{2}^2 }$

$F_{2} = \frac{9 * 10^{9} * 32 * 10^{-6} * 20 * 10^{-6} }{0.4^2 }\\F_{2} = 36 N$

Define the force magnitude on the +32 µC charge from the -60 µC charge as F₂ (the force on q₁ due to q₃).

$F_{3} = \frac{kq_{1}q_{3} }{x_{3}^2 }$

$F_{3} = -\frac{9 * 10^{9} * 32 * 10^{-6} * 60 * 10^{-6} }{0.6^2 }\\F_{3} =-48 N$

The resultant electrostatic force on the 32 µC charge is $F_{net} = |F_{2} + F_{3}|$

$F_{net} =| 36 + (-48)| \\F_{net} =|- 12 N| \\ F_{net} = 12 N$

7 0

1 day ago

You and your friend Peter are putting new shingles on a roof pitched at 20degrees . You're sitting on the very top of the roof w

Keith_Richards [1021]

Answer:

v₀ = 3.8 m/s

Explanation:

According to Newton's second law relating to the box:

∑F = m*a Formula (1)

∑F: the net force in Newton (N)

m: mass expressed in kilograms (kg)

a: acceleration measured in meters per second squared (m/s²)

Information known:

m = 2.1 kg, the mass of the box

d = 5.4m, the length of the roof

θ = 20° is the angle between the roof and the horizontal

μk = 0.51, the coefficient of kinetic friction between the box and the roof

g = 9.8 m/s², gravitational acceleration

Forces influencing the box:

The x-axis is oriented parallel to the box's movement on the roof, and the y-axis is oriented perpendicularly.

W: Weight of the box: directed vertically

N: Normal force: perpendicular to the roof's angle

fk: Frictional force: parallel to the direction along the roof

Calculating the weight of the box:

W = m*g = (2.1 kg)*(9.8 m/s²)= 20.58 N

x-y components of weight:

Wx= Wsin θ=(20.58)*sin(20)°=7.039 N

Wy= Wcos θ=(20.58)*cos(20)°= 19.34 N

Finding the Normal force:

∑Fy = m*ay ay = 0

N-Wy = 0

N=Wy = 19.34 N

Calculating the Friction force:

fk=μk*N= 0.51* 19.34 N = 9.86 N

We substitute into Formula (1) to determine the box's acceleration:

∑Fx = m*ax ax=a: acceleration of the box

Wx-fk = (2.1)*a

7.039 - 9.86 = (2.1)*a

-2.821 = (2.1)*a

a=(-2.821)/(2.1)

a = -1.34 m/s²

Considering the box's Kinematics:

Since the box undergoes uniformly accelerated motion, we use the following to find the final speed of the box:

vf² = v₀² + 2*a*d Formula (2)

Where:

d refers to displacement = 5.4 m

v₀ is the initial speed

vf represents the final speed = 0

a is the box's acceleration = -1.34 m/s²

Plugging in the values into Formula (2):

0² = v₀² + 2*(-1.34)*(5.4)

2*(1.34)*(5.4) = v₀²

$v_{o} =\sqrt{14.472}$

v₀ = 3.8 m/s

7 0

9 days ago

help please!! A runner at a speed of 3.5 m/s comes to a stop in 0.15 seconds. What is the deceleration?

Maru [1053]

Answer:

The runner's deceleration is -23.33 $\frac{m}{s^{2} }$

Given:

Initial speed = 3.5 $\frac{m}{s}$

Final speed = 0 $\frac{m}{s}$

Time taken = 0.15 s

To determine:

Deceleration of the runner =?

Used Formula:

Using the first equation of motion,

v = u + at

Where, v = final speed

u = initial speed

a = deceleration

t = duration

Solution:

<pusing the="" first="" equation="" of="" motion="">

v = u + at

Where, v = final speed

u = initial speed

a = deceleration

t = duration

0 = 3.5 + a (0.15)

-3.5 = 0.15 (a)

a = $\frac{-3.5}{0.15}$

a = -23.33 $\frac{m}{s^{2} }$

The negative sign indicates that this represents deceleration.

Hence, the deceleration of the runner is -23.33 $\frac{m}{s^{2} }$

</pusing>

7 0

1 day ago

A closed system of mass 10 kg undergoes a process during which there is energy transfer by work from the system of 0.147 kJ per

Softa [913]

Result: -50.005 kJ

Details:

Provided Data

mass of the system = 10 kg

work done = 0.147 kJ/kg

Elevation change $(\Delta h)=-50 m$

initial speed $(v_1)=15 m/s$

Final Speed $(v_2)=30 m/s$

Specific internal Energy $(\Delta U)=-5 kJ/kg$

according to the first Law of thermodynamics

$Q-W=\Delta H$

$\Delta H=\Delta KE+ \Delta PE +\Delta U$

where KE represents kinetic energy

PE indicates potential energy

U denotes internal Energy

$\Delta H=\frac{m(v_2^2-v_1^2)}{2}+mg(\Delta h)+\Delta U$

$Q=W+\Delta KE+ \Delta PE +\Delta U$

$Q=0.147\times 10+\frac{10\cdot (30^2-15^2)}{2}+10\cdot 9.81\cdot (-5)-5\times 10$

Q = 1.47 + 3.375 - 4.850 - 50

Q = -50.005 kJ

5 0

13 days ago