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2 months ago

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Your annoying little brother is dropping rocks out of his bedroom window on the 2nd floor. You are on the ground floor and watch

a rock fall past the patio door onto the ground below. In this problem, ignore air resistance. If the rock takes 0.28 s to cross the 2.3-m-high glass door, from what height above the top of the door was it dropped?

1 answer:

serg [3.5K]2 months ago

5 0

Answer:

Insufficient details provided; please clarify further.

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The weight of your car will also affect its _____.

inna [3103]

Stopping distance.

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1 month ago

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A 2 000-kg sailboat experiences an eastward force of 3 000 N by the ocean tide and a wind force against its sails with a magnitu

ValentinkaMS [3465]

Respuesta:

La magnitud de la aceleración resultante es 2.2 $m/s^2$

Explicación:

La masa (m) del velero es 2000 kg

La fuerza que actúa sobre el velero debido a la marea del océano es $F_1$ = 3000N

Hacia el este significa que se da a lo largo de la dirección positiva del eje x

Entonces $F_{1x}$ = 3000N y $F_{1y}$ = 0

La fuerza del viento que actúa sobre el velero es $F_2$ 6000N dirigida hacia el noroeste, lo que significa a un ángulo de 45 grados sobre el eje negativo x

Luego

$F_{2x}$ = -(6000N) cos 45 grados = -4242.6 N

$F_{2y}$ = (6000N) cos 45 grados = 4242.6 N

Por lo tanto, la fuerza neta que actúa sobre el velero en la dirección x es

$F_x = F_{1x}+ F_{2x}$

= - 3000 N + 4242.6 N

= - 3000 N +4242.6 N

= 1242.6N

La fuerza neta que actúa sobre el velero en la dirección y es

= 0+ 4242.6N

= 4242.6N $F_y = F_{1y}+ F_{2y}$ La magnitud de la fuerza resultante =

Usando el teorema de Pitágoras de 1243 N y 4243 N

4420.8 N $\sqrt{(1242.6)^2 + (4242.6)^2$ F = ma

$\sqrt{(1544054.76) + (17999654.8)}$

$\sqrt{(19543709.5)^2}$

= 2.2

4 0

14 days ago

One consequence of Einstein's theory of special relativity is that mass is a form of energy. This mass-energy relationship is pe

ValentinkaMS [3465]

Answer:

The energy unit is expressed as kg-m/s or Joules.

Explanation:

The relationship between mass and energy in physics is represented by:

$E=mc^2$

Where

m denotes the mass of the object

c signifies the speed of light

In the SI system, mass is measured in kilograms and the speed of light in m/s. Therefore, energy is defined in kg-m/s, which is equal to Joules.

Thus, the appropriate SI unit for energy is kg-m/s or Joules. This concludes the explanation.

6 0

1 month ago

Read 2 more answers

A skateboarder with mass ms = 54 kg is standing at the top of a ramp which is hy = 3.3 m above the ground. The skateboarder then

serg [3582]

Response:

A) $W_{ff} =-744.12J$

B) $F_f=-W_{ff}*sin\theta /hy = 112.75N$

C) $F_{f2}=207.58N$

Clarification:

The question is not fully provided. The complete question was:

A skateboarder with a mass of ms = 54 kg is at the top of a ramp with a height of hy = 3.3 m. He then jumps on his skateboard and goes down the ramp. His speed at the base is vf = 6.2 m/s.

Part (a) Formulate an expression for the work, Wf, done by the frictional force on the skateboarder in terms of the variables listed in the problem.

Part (b) The ramp is at an angle θ with the ground, where θ = 30°. Formulate an expression for the frictional force's magnitude, fr, between the skateboarder and the ramp.

Part (c) Upon reaching the bottom, the skateboarder continues with speed vf onto a grass-covered flat surface. The friction between the grass and the skateboarder brings him to a halt after 5.00 m. Determine the frictional force, Fgrass in newtons, between the skateboarder and the grass.

For part A), we assess the energy balance to determine the work done by the friction:

$W_{ff}=\Delta E$

$W_{ff}=1/2*m*vf^2-m*g*hy$

$W_{ff}=-744.12J$

For part B), we utilize the previously calculated work:

$W_{ff}=-F_f*(hy/sin\theta)$ Rearranging for friction force:

$F_f=-W_{ff}*sin\theta /hy$

$F_f=112.75N$

For part C), we first ascertain the acceleration through kinematic equations and subsequently find the frictional force using dynamic methods:

$Vf^2=Vo^2+2*a*d$

Rearranging for 'a':

$a=-3.844m/s^2$

Now, using dynamics:

$|F_f|=|m*a|$

$|F_f|=207.58N$

8 0

1 month ago

A charged wire of negligible thickness has length 2L units and has a linear charge density λ. Consider the electric field E-vect

Softa [3030]

Answer:

$E=2K\lambda d\dfrac{L }{d^2\sqrt{L^2+d^2}}$

Explanation:

Consider the following:

Length= 2L

Linear charge density=λ

Distance= d

K=1/(4πε)

The electric field measured at point P

$E=2K\int_{0}^{L}\dfrac{\lambda }{r^2}dx\ sin\theta$

$sin\theta =\dfrac{d}{\sqrt{d^2+x^2}}$

$r^2=d^2+x^2$

Thus,

$E=2K\lambda d\int_{0}^{L}\dfrac{dx }{(x^2+d^2)^{\frac{3}{2}}}$

Now, by applying integration to the equation above

$E=2K\lambda d\dfrac{L }{d^2\sqrt{L^2+d^2}}$

4 0

1 month ago