•What is the gravitational potential energy of a girl

A Ford Mustang weighs about 3500 pounds, and can accelerate from 0-60 MPH in about 5 seconds. What force is responsible for this

Hunter-Best [27]

We will apply the concepts related to Newton's second law. At the same time we will convert everything to the system of international units.

$m = 3500lb = 1587.57kg$

The values of the velocities are,

$\text{Initial Velocity} = V_i = 0$

$\text{Final Velocity} = V_f = 60mph = 26.822m/s$

We know that the acceleration is equivalent to the change of the speed in a certain time therefore

$a = \frac{v_f-v_i}{t}$

$a = \frac{26.822-0}{5}$

$a = 5.36m/s^2$

Now applying the Newton's second law we have,

$F= ma$

$F = (1587.57)(5.36)$

$F = 8516.36N$

Therefore the approximate magnitude is 8516.36N

5 0

4 years ago

A concrete slab shown in Figure 5 is being lifted by using three cables connected to the slab at points A, B and C. The slab is

xxMikexx [17]

Answer:

Fad = 28.8 kN

Fbd = 16.4 kN

Fcd = 28.1 kN

Explanation:

First, find the length of each cable.

AD = √((2 m)² + (0.5 m)² + (2.5 m)²)

AD = √10.5 m

AD ≈ 3.24 m

BD = √((1.5 m)² + (1 m)² + (2.5 m)²)

BD = √9.5 m

BD ≈ 3.08 m

CD = √((1 m)² + (1 m)² + (2.5 m)²)

CD = √8.25 m

CD ≈ 2.87 m

Next, use similar triangles to find the x, y, and z components of each tension force.

Fadx = 2/3.24 Fad = 0.617 Fad

Fady = 0.5/3.24 Fad = 0.154 Fad

Fadz = 2.5/3.24 Fad = 0.772 Fad

Fbdx = 1.5/3.08 Fbd = 0.487 Fbd

Fbdy = 1/3.08 Fbd = 0.324 Fbd

Fbdz = 2.5 / 3.08 Fbd = 0.811 Fbd

Fcdx = 1/2.87 Fcd = 0.348 Fcd

Fcdy = 1/2.87 Fcd = 0.348 Fcd

Fcdz = 2.5/2.87 Fcd = 0.870 Fcd

Now sum the forces in the x, y, and z directions:

∑Fx = ma

-0.617 Fad + 0.487 Fbd + 0.348 Fcd = 0

∑Fy = ma

-0.154 Fad − 0.324 Fbd + 0.348 Fcd = 0

∑Fz = ma

60 kN − 0.772 Fad − 0.811 Fbd − 0.870 Fcd = 0

To solve this system of equations algebraically, start by subtracting the first two equations, eliminating Fcd.

-0.463 Fad + 0.811 Fbd = 0

0.811 Fbd = 0.463 Fad

Fbd = 0.571 Fad

Substitute into either of the first two equations:

-0.617 Fad + 0.487 (0.571 Fad) + 0.348 Fcd = 0

-0.617 Fad + 0.278 Fad + 0.348 Fcd = 0

-0.339 Fad + 0.348 Fcd = 0

0.348 Fcd = 0.339 Fad

Fcd = 0.975 Fad

Now substituting into the third equation:

60 kN − 0.772 Fad − 0.811 Fbd − 0.870 Fcd = 0

60 kN − 0.772 Fad − 0.811 (0.571 Fad) − 0.870 (0.975 Fad) = 0

60 kN − 0.772 Fad − 0.463 Fad − 0.849 Fad = 0

60 kN − 2.083 Fad = 0

Fad = 28.8 kN

Solving for the other two tension forces:

Fbd = 0.571 Fad = 16.4 kN

Fcd = 0.975 Fad = 28.1 kN

3 0

3 years ago

Read 2 more answers

A student is building a simple circuit with a battery, light bulb, and copper wires. When she connects the wires to the battery

denis23 [38]

It’s will be B because the circuit had a open or close so if that doesn’t work than it’s because it’s open

5 0

4 years ago

Read 2 more answers

How can a small human retina detect objects larger than itself?

muminat

The correct answer would be A) the lenses of the eye change the size of images

5 0

3 years ago

Calculate the electric field at the center of a square

pantera1 [17]

Answer:

$E_y=1175510.2\ N.C^{-1}$

The Magnitude of electric field is in the upward direction as shown directly towards the charge $q_1$ .

Explanation:

Given:

side of a square, $a=52.5\ cm$

charge on one corner of the square, $q_1=+45\times 10^{-6}\ C$

charge on the remaining 3 corners of the square, $q_2=q_3=q_4=-27\times 10^{-6}\ C$

Distance of the center from each corners $=\frac{1}{2} \times diagonals$

$diagonal=\sqrt{52.5^2+52.5^2}$

$diagonal=74.25\cm=0.7425\ m$

∴Distance of center from corners, $b=0.3712\ m$

Now, electric field due to charges is given as:

$E=\frac{1}{4\pi\epsilon_0}\times \frac{q}{b^2}$

For charge $q_1$ we have the field lines emerging out of the charge since it is positively charged:

$E_1=9\times 10^9\times \frac{45\times 10^{-6}}{0.3712^2}$

$E_1=2938775.5\ N.C^{-1}$

Force by each of the charges at the remaining corners:

$E_2=E_3=E_4=9\times 10^9\times \frac{27\times 10^{-6}}{0.3712^2}$

$E_2=E_3=E_4=1763265.3\ N.C^{-1}$

Now, net electric field in the vertical direction:

$E_y=E_1-E_4$

$E_y=1175510.2\ N.C^{-1}$

Now, net electric field in the horizontal direction:

$E_y=E_2-E_3$

$E_y=0\ N.C^{-1}$

So the Magnitude of electric field is in the upward direction as shown directly towards the charge $q_1$ .

8 0

3 years ago