All categories

3 years ago

What two forces act on a falling object ?

Physics

1 answer:

goldenfox [79]3 years ago

5 0

Answer:

The two forces acting on the object are weight due to gravity pulling the object towards earth, and drag resisting this motion. When the object is first released, drag is small as velocity is low, so the resultant force is down. This means the object accelerates towards earth.

You might be interested in

Par 1/2

BaLLatris [955]

part 1

mass = ρ x V

mass = 1739 kg/m³ x 3.8 km³ = 6608.2 kg

PE (potential energy)= mgh

PE = 6608.2 kg x 9.81 x 403

PE = 2.61 x 10⁷ J

part 2

megaton of TNT (Mt) =4.2 x 10¹⁵ J

convert PE to Mt:

2.61 x 10⁷ J : 4.2 x 10¹⁵ J = 6.21 x 10⁻⁹ Mt

4 0

3 years ago

Mrs. Paul and Dr. Mykannen ride a tandem bike on the beach. They ride along the beach for about 455 meters. Their final velocity

algol [13]

Answer:

Time = 80.91 seconds

Explanation:

Given the following data;

Velocity = 5.50 m/s.

Distance = 445 meters

To find the time;

Velocity can be defined as the rate of change in displacement (distance) with time. Velocity is a vector quantity and as such it has both magnitude and direction.

Mathematically, velocity is given by the equation;

$Velocity = \frac{distance}{time}$

Substituting into the formula, we have;

5.5 = 445/time

Time = 445/5.5

Time = 80.91 seconds

5 0

3 years ago

Two ice skaters, each with a mass of 72.0 kg, are skating at 5.45 m/s when they collide and stick together. If the angle between

Rudiy27

Answer:

The skater 1 and skater 2 have a final speed of 2.02m/s and 2.63m/s respectively.

Explanation:

To solve the problem it is necessary to go back to the theory of conservation of momentum, specifically in relation to the collision of bodies. In this case both have different addresses, consideration that will be understood later.

By definition it is known that the conservation of the moment is given by:

$m_1v_1+m_2v_2=(m_1+m_2)v_f$

Our values are given by,

$m_1=m_2=72Kg$

As the skater 1 run in x direction, there is not component in Y direction. Then,

Skate 1:

$v_{x1}=5.45m/s$

$v_{y1}=0$

Skate 2:

$v_{x2} = 5.45*cos105= -1.41m/s$

$v_{y2} = 5.45*sin105 = 5.26m/s$

Then, if we applying the formula in X direction:

m_1v_{x1}+m_2v_{x2}=(m_1+m_2)v_{fx}

75*5.45-75*1.41=(75+75)v_{fx}

Re-arrange and solving for v_{fx}

v_{fx}=\frac{4.04}{2}

v_{fx}=2.02m/s

Now applying the formula in Y direction:

$m_1v_{y1}+m_2v_{y2}=(m_1+m_2)v_{fy}$

$0+75*5.25=(75+75)v_{fy}$

$v_{fy}=\frac{5.25}{2}$

$v_{fy}=2.63m/s$

Therefore the skater 1 and skater 2 have a final speed of 2.02m/s and 2.63m/s respectively.

6 0

3 years ago

What is the speed of sound in air that has a temperature of 19.9 celcius

klasskru [66]

Answer4.4 :

Explanation:

7 0

3 years ago

An object is originally moving at a constant velocity of 8 m/s in the -x direction. It moves at this constant velocity for 3 sec

aivan3 [116]

Answer:

244.64m

Explanation:

First, we find the distance traveled with constant velocity. It's simply multiplying velocity time the time that elapsed:

$x = V*t = -8\frac{m}{s} *3s = -24m$

After this, the ball will start traveling with a constant acceleration motion. Due to the fact that the acceleration is the opposite direction to the initial velocity, this motion will have 2 phases:

1. The velocity will start to decrease untill it reaches 0m/s.

2. Then, the velocity will start to increase at the rate of the acceleration.

The distance that the ball travels in the first phase can be found with the following expression:

$v^2 = v_0^2 + 2a*d$

Where v is the final velocity (0m/s), v_0 is the initial velocity (-8m/s) and a is the acceleration (+9m/s^2). We solve for d:

$d = \frac{v^2 - v_0^2}{2a} = \frac{(0m/s)^2 - (-8m/s)^2}{2*7m/s^2}= -4.57m$

Now, before finding the distance traveled in the second phase, we need to find the time that took for the velocity to reach 0:

$t_1 = \frac{v}{a} = \frac{8m/s}{7m/s^2} = 1.143 s$

Then, the time of the second phase will be:

$t_2 = 9s - t_1 = 9s - 1.143s = 7.857s$

Using this, we using the equations for constant acceleration motion in order to calculate the distance traveled in the second phase:

$x = \frac{1}{2}a*t^2 + v_0*t + x_0$

V_0, the initial velocity of the second phase, will be 0 as previously mentioned. X_0, the initial position, will be 0, for simplicity:

$x = \frac{1}{2}*7\frac{m}{s^2}*t^2 + 0m/s*t + 0m = 216.07m$

So, the total distance covered by this object in meters will be the sum of all the distances we found:

$x_total = 24m + 4.57m + 216.07m = 244.64m$

8 0

3 years ago