How many joules of work are done on a box when a force of 25 N pushes it 3 m?

How far can a sound wave travel in 90 seconds when the ambient air temperature is 10 C?

Ksju [112]

Answer:

s = 30330.7 m = 30.33 km

Explanation:

First we need to calculate the speed of sound at the given temperature. For this purpose we use the following formula:

v = v₀√[T/273 k]

where,

v = speed of sound at given temperature = ?

v₀ = speed of sound at 0°C = 331 m/s

T = Given Temperature = 10°C + 273 = 283 k

Therefore,

v = (331 m/s)√[283 k/273 k]

v = 337 m/s

Now, we use the following formula to calculate the distance traveled by sound:

s = vt

where,

s = distance traveled = ?

t = time taken = 90 s

Therefore,

s = (337 m/s)(90 s)

6 0

3 years ago

An 84.0 kg sprinter starts a race with an acceleration of 1.76 m/s2. If the sprinter accelerates at that rate for 11 m, and then

gulaghasi [49]

Answer:

t=17.838s

Explanation:

The displacement is divided in two sections, the first is a section with constant acceleration, and the second one with constant velocity. Let's consider the first:

The acceleration is, by definition:

$a=\frac{dv}{dt}=1.76$

So, the velocity can be obtained by integrating this expression:

$v=1.76t$

The velocity is, by definition: $v=\frac{dx}{dt}$ , so

$dx=1.76tdt\\x=1.76\frac{t^{2}}{2}$ .

Do x=11 in order to find the time spent.

$11=1.76\frac{t^2}{2}\\ t^2=\frac{2*11}{76} \\t=\sqrt{12.5}=3.5355s$

At this time the velocity is: $v=1.76t=1.76*3.5355s=6.2225\frac{m}{s}$

This velocity remains constant in the section 2, so for that section the movement equation is:

$x=v*t\\t=\frac{x}{v}$

The left distance is 89 meters, and the velocity is $6.2225\frac{m}{s}$ , so:

$t=\frac{89}{6.2225}=14.303s$

So, the total time is 14.303+3.5355s=17.838s

7 0

3 years ago

PLEASE HELP!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! IM DESPERATE!!!!!!!!!!!!!! EVEN IF YOU DONT KNOW , GIVE A GUESS. What would

dezoksy [38]

Well, A = T or U C = G G = C T or U = A So it would be like this; DNA Sequence: GCTAATTGCATCCGA The Complementary Sequence: CGATTAACGTAGGCT Hope this helped :)

3 0

3 years ago

System A has masses m and m separated by a distance r; system B has masses m and 2m separated by a distance 2r; system C has mas

Anna [14]

Answer:

System D --> System C --> System A --> System B

Explanation:

The gravitational force between two masses m1, m2 separated by a distance r is given by:

$F=G \frac{m_1 m_2}{r^2}$

where G is the gravitational constant. Let's apply this formula to each case now to calculate the relative force for each system:

System A has masses m and m separated by a distance r:

$F=G\frac{m \cdot m}{r^2}=G \frac{m^2}{r^2}$

system B has masses m and 2m separated by a distance 2r:

$F=G\frac{m \cdot 2m}{(2r)^2}=G \frac{2m^2}{4r^2}=\frac{1}{2} G \frac{m^2}{r^2}$

system C has masses 2m and 3m separated by a distance 2r:

$F=G\frac{2m \cdot 3m}{(2r)^2}=G \frac{6m^2}{4r^2}=\frac{3}{2} G \frac{m^2}{r^2}$

system D has masses 4m and 5m separated by a distance 3r:

$F=G\frac{4m \cdot 5m}{(3r)^2}=G \frac{20m^2}{9r^2}=\frac{20}{9} G \frac{m^2}{r^2}$

Now, by looking at the 4 different forces, we can rank them from the greatest to the smallest force, and we find:

System D --> System C --> System A --> System B

5 0

3 years ago

A 5.50 kg sled is initially at rest on a frictionless horizontal road. The sled is pulled a distance of 3.20 m by a force of 25.

kiruha [24]

(a) 69.3 J

The work done by the applied force is given by:

$W=Fd cos \theta$

where:

F = 25.0 N is the magnitude of the applied force

d = 3.20 m is the displacement of the sled

$\theta=30^{\circ}$ is the angle between the direction of the force and the displacement of the sled

Substituting numbers into the formula, we find

$W=(25.0 N)(3.20 m)(cos 30^{\circ})=69.3 J$

(b) 0

The problem says that the surface is frictionless: this means that no friction is acting on the sled, therefore the energy dissipated by friction must be zero.

(c) 69.3 J

According to the work-energy theorem, the work done by the applied force is equal to the change in kinetic energy of the sled:

$\Delta K = W$

where

$\Delta K$ is the change in kinetic energy

W is the work done

Since we already calculated W in part (a):

W = 69.3 J

We therefore know that the change in kinetic energy of the sled is equal to this value:

$\Delta K=69.3 J$

(d) 4.9 m/s

The change in kinetic energy of the sled can be rewritten as:

$\Delta K=K_f - K_i = \frac{1}{2}mv^2-\frac{1}{2}mu^2$ (1)

where

Kf is the final kinetic energy

Ki is the initial kinetic energy

m = 5.50 kg is the mass of the sled

u = 0 is the initial speed of the sled

v = ? is the final speed of the sled

We can calculate the variation of kinetic energy of the sled, $\Delta K$ , after it has travelled for d=3 m. Using the work-energy theorem again, we find

$\Delta K= W = Fd cos \theta =(25.0 N)(3.0 m)(cos 30^{\circ})=65.0 J$

And substituting into (1) and re-arrangin the equation, we find

$v=\sqrt{\frac{2 \Delta K}{m}}=\sqrt{\frac{2(65.0 J)}{5.50 kg}}=4.9 m/s$

6 0

3 years ago