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topjm [15]
3 years ago
15

An object is released from rest at time t = 0 and falls through the air, which exerts a resistive force such that the accelerati

on a of the object is given by a = g – bV, where V is the object's speed and b is a constant. If limiting cases for large and small values of t are considered, which of the following is a possible expression for the speed of the object as an explicit function of time?
a. v = g(l - e-bt)/b.
b. v = (gebt)/b.
c. v = gt - bt2.
d. v = (g + a)t/b.
e. v = V0 + gt, v0.
Physics
1 answer:
lubasha [3.4K]3 years ago
6 0

Answer:

The correct option is

a. v = g (1-e^{-bt})/b

Explanation:

Time at which the object start fall t = 0

The acceleration a is given by a = g - bV

Where V = Speed of the object

Speed V² = u² + 2·a·h

However with the drag force the object will approach terminal velocity as t becomes progressively larger whereby v will stop increasing

Option a. is the only option that has  limiting value of v which is in the range of g as t increases ∴ option a. is the correct option.

v = g (1-e^{-bt})/b  as t increases (1-e^{-bt}) → 1 s and v→ g/b m/s

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radar wave is transmitted and later reflected off an aircraft and recieved 1.4×10^3 sec after being sent out. how far is the air
lutik1710 [3]

I think there's a typo because the answer I'm getting is very large.

This is what I'm getting

--------------------------------------

c = speed of light

c = 3.0 x 10^8 m/sec approximately

This is roughly 300 million meters per second

The time it takes the signal to reach the aircraft and come back is 1.4 x 10^3 seconds. Half of this time period is going one direction (say from the radar station to the aircraft), so (1.4 x 10^3)/2 = 7.0 x 10^2 seconds is spent going in this one direction.

distance = rate*time

d = r*t

d = (3.0 x 10^8) * (7.0 x 10^2)

d = (3.0*7.0) x (10^8*10^2)

d = 21.0 x 10^(8+2)

d = 21.0 x 10^10

d = (2.1 x 10^1) * 10^10

d = 2.1 x (10^1*10^10)

d = 2.1 x 10^11 meters

d = 210,000,000,000 meters (this is 210 billion meters; equivalent to roughly 130,487,950 miles)

3 0
3 years ago
A 20 kg wagon is pulled along the level ground by a rope inclined at 30 degree above the horizontal. A friction force of 30 N op
Elan Coil [88]

(a) 34.6 N

To solve the problem, we have to analyze the forces acting along the horizontal direction.

We have:

- Forward: the component of the pull parallel to the ground, which is

F cos \theta

where

F is the magnitude of the pull

\theta=30^{\circ} is the angle

- Backward: the force of friction, which is

F_f = 30 N

So, the equation of motion is

F cos \theta - F_f = ma

where

m = 20 kg is the mass of the wagon

a is the acceleration

In this part, the wagon is moving at constant speed, so a =0 and the equation becomes

F cos \theta - F_f = 0

Therefore, we can find the pulling force:

F=\frac{F_f}{cos \theta}=\frac{30}{cos 30}=34.6 N

(b) 43.9 N

In this case, the acceleration is

a=0.40 m/s^2

So, the equation of motion in this case is

F cos \theta - F_f = ma

So this time we have to take into account the term (ma).

Using the  same data as before:

m = 20 kg

\theta=30^{\circ}

F_f = 30 N

We find the new magnitude of F:

F=\frac{ma+F_f}{cos \theta}=\frac{(20)(0.40)+30}{cos 30}=43.9 N

6 0
3 years ago
A person pushes horizontally with a force of 220. N on a 61.0 kg crate to move it across a level floor. The coefficient of kinet
Ede4ka [16]

Answer:

(a) 161.57 N

(b) 0.958 m/s^2

Explanation:

Force applied, F = 220 N

mass of crate, m = 61 kg

μ = 0.27

(a) The magnitude of the frictional force,

f = μ N

where, N is the normal reaction

N = m x g = 61 x 9.81 = 598.41 N

So, the frictional force, f = 0.27 x 598.41

f = 161.57 N

(b) Let a be the acceleration of the crate.

Fnet = F - f = 220 - 161.57

Fnet = 58.43 N

According to newton's second law

Fnet = mass x acceleration

58.43 = 61 x a

a = 0.958 m/s^2

Thus, the acceleration of the crate is 0.958 m/s^2.  

7 0
3 years ago
When heat is removed from a substance , describe how the molecules are affected
kkurt [141]

I believe they would electron rate would slow down and the molecules would shrink.

I am almost positive that this is correct. I hope it helps!

6 0
3 years ago
Read 2 more answers
Newton’s empirical law of cooling/warming of an object is given by ( ), T Tm k dt dT = − where k is a constant of proportionalit
pychu [463]

Answer:

The time for the cake to cool off to room temperature is

approximately 30 minutes.

Let T_{0} = 70^{0}F be the temperature and T that of the body

Explanation:

 Our Tm = 70, the initial-value problem is

\frac{DT}{dt} = <em>k</em>(T − 70), T(0) = 300

Solving the equation, we get

\frac{DT}{t-70} = <em>kdt</em>

In [T-70]= <em>kt </em>+C_{1}

    T   =  70  + C_{2} e^{kt}

Finding he value for C_{2} using the initial value of T (0)= 300, therefore we get:

300=70+C_{2}

C_{2} = 230 therefore

T= 70+ 230 e^{kt}

Finding the value for <em>k </em>using T (3)  = 200, therefore we get

T (3) = 200

e^{3k} = \frac{13}{23}

<em>K </em>= \frac{1}{3} in \frac{13}{23}

= -0.19018

Therefore

T(t) = 70+230e^{-0.19018}

4 0
3 years ago
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