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Romashka-Z-Leto [24]
3 years ago
7

A small piece of Styrofoam packing material is dropped from a height of 2.60 m above the ground. Until it reaches terminal speed

, the magnitude of its acceleration is given by a = g − Bv. After falling 0.600 m, the Styrofoam effectively reaches terminal speed, and then takes 4.50 s more to reach the ground.
(a) What is the value of the constant B? s-1.
(b) What is the acceleration at t = 0? [m/s2 (down)].
(c) What is the acceleration when the speed is 0.150 m/s? [m/s2 (down)].
Physics
1 answer:
Vadim26 [7]3 years ago
7 0

Answer:

Explanation:

a ) After the attainment of terminal speed , object takes 4.5 s to cover a distance of 2 m

So terminal speed V = 2 / 4.5

= .444 m /s

When it attains terminal speed , acceleration becomes zero

0 = g - B x .444

B = 22.25 s⁻¹

b ) At t = 0 , v = 0

a = g - B v

a = g at t = 0

c ) When v = .15

a = g - 22.25 x .15

= 9.8 - 3.31

= 6.5  m /s²

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As a capacitor is being charged in an RC circuit. the current flowing through a resistor isa) increasingb) decreasingc) constant
Montano1993 [528]
<h2>Answer: decreasing</h2>

An RC circuit is an electrical circuit composed of resistors and capacitors, where the charging time T of the circuit is proportional to the magnitude of the electrical resistance R and the capacity C of the capacitor.  

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T=R.C

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<h2>When a capacitor is being charged in an RC circuit, the current flowing through a resistor <u>decreases</u>.</h2>

And the correct option is b.

4 0
3 years ago
A ball is tossed with enough speed straight up so that it is in the air several seconds. (a) What is the velocity of the ball wh
irina1246 [14]

(a) Zero

When the ball reaches its highest point, the direction of motion of the ball reverses (from upward to downward). This means that the velocity is changing sign: this also means that at that moment, the velocity must be zero.

This can be also understood in terms of conservation of energy: when the ball is tossed up, initially it has kinetic energy

K=\frac{1}{2}mv^2

where m is the ball's mass and v is the initial speed. As it goes up, this kinetic energy is converted into potential energy, and when the ball reaches the highest point, all the kinetic energy has been converted into potential energy:

U=mgh

where g is the gravitational acceleration and h is the height of the ball at highest point. At that point, therefore, the potential energy is maximum, while the kinetic energy is zero, and so the velocity is also zero.

(b) 9.8 m/s upward

We can find the velocity of the ball 1 s before reaching its highest point by using the equation:

a=\frac{v-u}{t}

where

a = g = -9.8 m/s^2 is the acceleration due to gravity, which is negative since it points downward

v = 0 is the final velocity (at the highest point)

u is the initial velocity

t = 1 s is the time interval

Solving for u, we find

u=v-at = 0 -(-9.8 m/s^2)(1 s)= +9.8 m/s

and the positive sign means it points upward.

(c) -9.8 m/s

The change in velocity during the 1-s interval is given by

\Delta v = v -u

where

v = 0 is the final velocity (at the highest point)

u = 9.8 m/s is the initial velocity

Substituting, we find

\Delta v = 0 - (+9.8 m/s)=-9.8 m/s

(d) 9.8 m/s downward

We can find the velocity of the ball 1 s after reaching its highest point by using again the equation:

a=\frac{v-u}{t}

where this time we have

a = g = -9.8 m/s^2 is the acceleration due to gravity, still negative

v  is the final velocity (1 s after reaching the highest point)

u = 0 is the initial velocity (at the highest point)

t = 1 s is the time interval

Solving for v, we find

v = u+at = 0 +(-9.8 m/s^2)(1 s)= -9.8 m/s

and the negative sign means it points downward.

(e) -9.8 m/s

The change in velocity during the 1-s interval is given by

\Delta v = v -u

where here we have

v = -9.8 m/s is the final velocity (1 s after reaching the highest point)

u = 0 is the initial velocity (at the highest point)

Substituting, we find

\Delta v = -9.8 m/s - 0=-9.8 m/s

(f) -19.6 m/s

The change in velocity during the overall 2-s interval is given by

\Delta v = v -u

where in this case we have:

v = -9.8 m/s is the final velocity (1 s after reaching the highest point)

u = +9.8 m/s is the initial velocity (1 s before reaching the highest point)

Substituting, we find

\Delta v = -9.8 m/s - (+9.8 m/s)=-19.6 m/s

(g) -9.8 m/s^2

There is always one force acting on the ball during the motion: the force of gravity, which is given by

F=mg

where

m is the mass of the ball

g = -9.8 m/s^2 is the acceleration due to gravity

According to Newton's second law, the resultant of the forces acting on the body is equal to the product of mass and acceleration (a), so

mg = ma

which means that the acceleration is

a= g = -9.8 m/s^2

and the negative sign means it points downward.

7 0
3 years ago
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