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3 years ago

1. Your big brother, despite warning, drops a penny off the observation

Physics

1 answer:

Tpy6a [65]3 years ago

4 0

Answer:

The marble will take 8.813 seconds to hit the ground.

Explanation:

The marble will fall according to this motion equation: y(final) = y(initial) + v(initial)×t + 1/2×a×t×t

y(final)=0 m, the marble's final position in relation to the ground

y(initial)=381 m, the marble's initial position in relation to the ground

v(initial)=0 m/s (meters per second), the marble's initial velocity (speed with direction) when fall first begins

t=? seconds, this is the amount of time that passes while falling, the variable in this problem

a=-9.81 m/s/s, the rate at which the velocity of the marble changes, acceleration. On Earth, neglecting friction, acceleration is a constant at negative 9.81 meters per second squared

Plug the values into the equation: 0= 381 + 0t + 1/2(-9.81)(t)(t)

Subtract 381 from both sides and disregard 0t because it has no value and no longer matters: -381= 1/2(-9.81)(t)(t)

Multiply out the equation: -381= -4.905(t)(t)

Divide both sides by -4.905 to isolate the variable: 77.676=(t)(t)

Find the square root of both sides: t=8.813 seconds

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A spotlight on the ground is shining on a wall 24m away. If a woman 2m tall walks from the spotlight toward the building at a sp

Lubov Fominskaja [6]

Answer:

$\dfrac{dy}{dt}=-0.059\ m/s$

Explanation:

It is given that,

Distance between the spotlight and the wall, y = 24 m

Height of the woman, h = 2 m

The woman walks toward the building at the rate of 0.6 m/s, $\dfrac{dx}{dt}=0.6\ m/s$

In the attached figure, triangle ABC and MNC are similar. So,

$\dfrac{2}{y}=\dfrac{x}{24}$ ............(1)

$y=\dfrac{48}{x}$

When she is 2 meters from the building. So x = 24-2 = 22 m

$y=\dfrac{48}{22}=2.18\ m$

Differentiating equation (1) i.e.

$xy=48$

$x.\dfrac{dy}{dt}+y.\dfrac{dx}{dt}=0$

$22.\dfrac{dy}{dt}+2.18\times 0.6=0$

$\dfrac{dy}{dt}=-0.059\ m/s$

So, her shadow is decreasing at the rate of 0.059 m/s. Hence, this is the required solution.

7 0

4 years ago

what is the average speed of (a) a car that travels 400m in 20s. and (b) an athlete who runs 1500m in 4 minutes

zmey [24]

Answer:

a) 20 m/s

b) 37.5 m)s

Explanation:

Average speed = total distance ÷ total time

=> (a) average speed of a car that travels 400m in 20s

= 400/20 = 20 m/s

& (b) average speed of an athlete who runs 1500m in 4 minutes (or 4×60=240 seconds)

= 1500/240 = 37.5 m/s

5 0

3 years ago

Why did she use an red infra lamp

mixas84 [53]

To act as the Sun' was accepted but if you put 'sunlight' alone it was not accepted. The examiner wanted you to state that the infra red radiation was needed to warm up the water.

4 0

3 years ago

A box with mass m is dragged across a level floor with coefficient of kinetic friction μk by a rope that is pulled upward at an

Komok [63]

Answer:

a) F = μk mg Cosθ

b) F = 279.78 N

Explanation:

a) F = μk R

Based on the description in the question, the horizontal reaction is:

R = mg Cosθ

The force required to move the box with constant speed in terms of m, μk, θ, and g is :

F = μk mg Cosθ

b) If m = 90 kg

g = 9.8 m/s²

μk=0.35

θ = 25⁰

Force required to slide the 90-kg patient across a floor at constant speed by pulling on him at an angle of 25∘ above the horizontal will be:

F = μk mg Cosθ

F = 0.35 * 90 * 9.8 * cos25

F = 279.78 N

5 0

4 years ago

A constant torque is applied to a rigid wheel whose moment of inertia is 2.0 kg · m2 around the axis of rotation. If the wheel s

Jlenok [28]

Answer:

The applied torque is 3.84 N-m.

Explanation:

Given that,

Moment of inertia of the wheel is $2\ kg-m^2$

Initial speed of the wheel is 0 (at rest)

Final angular speed is 25 rad/s

Time, t = 13 s

The relation between moment of inertia and torque is given by :

$\tau=I\alpha \\\\\tau=I\times \dfrac{\omega_f}{t}\\\\\tau=2\times \dfrac{25}{13}\\\\\tau=+3.84\ N-m$

So, the applied torque is 3.84 N-m.

4 0

3 years ago