A 3.0-kilogram cart possesses 96 joules of kinetic energy. Calculate the speed of the car.

A 215-kg load is hung on a wire of length of 3.60 m, cross-sectional area 2.00 10-5 m2, and Young's modulus 8.00 1010 N/m2. What

Y_Kistochka [10]

Answer:

4.74 * 10^-3

Explanation:

6 0

3 years ago

Ferdinand the frog is hopping from lily pad to lily pad in search of a good fly

loris [4]

Answer: $36.86\°$

Explanation:

According to the described situation we have the following data:

Horizontal distance between lily pads: $d=2.4 m$

Ferdinand's initial velocity: $V_{o}=5 m/s$

Time it takes a jump: $t=0.6 s$

We need to find the angle $\theta$ at which Ferdinand jumps.

In order to do this, we first have to find the horizontal component (or x-component) of this initial velocity. Since we are dealing with parabolic movement, where velocity has x-component and y-component, and in this case we will choose the x-component to find the angle:

$V_{x}=\frac{d}{t}$ (1)

$V_{x}=\frac{2.4 m}{0.6 s}$ (2)

$V_{x}=4 m/s$ (3)

On the other hand, the x-component of the velocity is expressed as:

$V_{x}=V_{o}cos\theta$ (4)

Substituting (3) in (4):

$4 m/s=5 m/s cos\theta$ (5)

Clearing $\theta$ :

$\theta=cos^{-1} (\frac{4 m/s}{5 m/s})$

$\theta=36.86\°$ This is the angle at which Ferdinand the frog jumps between lily pads

4 0

3 years ago

Three boxes in contact rest side-by-side on a smooth, horizontal floor. Their masses are 5.0-kg, 3.0-kg, and 2.0-kg, with the 3.

Ivahew [28]

Answer:

(a)Look at the attached graphic

(b)

(b)-1 Equation 1 : m1= 5kg

50-F1= 5 *a

(b)-2 Equation 2 : m2= 3kg

F1-F2= 3 *a

(b)-3 Equation 3 : m3= 2kg

F2 = 2*a

(c) F1 =25 N

(d) F2 =10 N

Explanation:

We apply Newton's second law:

∑F = m*a (Formula 1)

∑F : algebraic sum of the forces in Newton (N)

m : mass in kilograms (kg)

a : acceleration in meters over second square (m/s²)

(a) Draw the free-body diagrams for each of the boxes

Look at the attached graphic

(b) Write Newton’s equation for each mass along the horizontal direction.

Data: m1= 5.0-kg ,m2= 3.0-kg , ,m3= 2.0-kg

Look m1 free-body diagram:

∑Fx = m1*a

50-F1= 5 *a Equation 1

Look m2 free-body diagram:

∑Fx = m2*a

F1-F2= 3 *a Equation 2

Look m3 free-body diagram:

∑Fx = m3*a

F2 = 2*a Equation 3

(c) What magnitude force does the 3.0-kg box exert on the 5.0- kg box?

Look Free body diagram of the mass set

∑Fx = m*a m= m1+m2+m3= 5+3+2 = 10 kg

50 = 10*a

a= 50/10 = 5 m/s²

We replace a = 5 m/s² in the equation 1:

50-F1= 5 *5

50-25= F1

F1 = 25 N

(d) What magnitude force does the 3.0-kg box exert on the 2.0kg box?

We replace a= 5 m/s² in the equation 3

F2 = 2*5 = 10 N

4 0

3 years ago

What is the net force in the following diagram?

ValentinkaMS [17]

I don’t see a diagram

5 0

2 years ago

A train whistle is heard at 300 Hz as the train approaches town. The train cuts its speed in half as it nears the station, and t

givi [52]

To solve this problem we will apply the concepts related to the Doppler effect. The Doppler effect is the change in the perceived frequency of any wave movement when the emitter, or focus of waves, and the receiver, or observer, move relative to each other. Mathematically it can be described as,

$f = f_0 (\frac{v_0}{v_0-v})$