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1 year ago

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A 1500 kg car is pushing a 4000 kg truck. The car and truck are accelerating at 2.0 m/s^2. Assuming that the frictional force on

the truck is negligible, what force is the truck exerting on the car?

1 answer:

Crazy boy [7]1 year ago

7 0

To solve this problem we will use the Force equation according to the definition given in Newton's second law. There we have that the Force is equal to

$F =ma$

Where,

m = mass

a = acceleration

Our values are given as

$m_1 = 1500 kg$

$m_2 = 4000 kg$

$a = 2.0 m/s^2$

Considering that both mass are equal to one, we have that:

$F = (m_1+m_2)* a$

$F = (1500 + 4000)(2.0)$

$F = 11000 N = 11kN$

Therefore the truck exert a force on the car of 11kN

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A section of highway has the following flowdensity relationship q = 50k − 0.156k2 [with q in veh/h and k in veh/mi]. What is the

lions [1.4K]

Answer:

a) capacity of the highway section = 4006.4 veh/h

b) The speed at capacity = 25 mph

c) The density when the highway is at one-quarter of its capacity = k = 21.5 veh/mi or 299 veh/mi

Explanation:

q = 50k - 0.156k²

with q in veh/h and k in veh/mi

a) capacity of the highway section

To obtain the capacity of the highway section, we first find the k thay corresponds to the maximum q.

q = 50k - 0.156k²

At maximum flow density, (dq/dk) = 0

(dq/dt) = 50 - 0.312k = 0

k = (50/0.312) = 160.3 ≈ 160 veh/mi

q = 50k - 0.156k²

q = 50(160.3) - 0.156(160.3)²

q = 4006.4 veh/h

b) The speed at the capacity

U = (q/k) = (4006.4/160.3) = 25 mph

c) the density when the highway is at one-quarter of its capacity?

Capacity = 4006.4

One-quarter of the capacity = 1001.6 veh/h

1001.6 = 50k - 0.156k²

0.156k² - 50k + 1001.6 = 0

Solving the quadratic equation

k = 21.5 veh/mi or 299 veh/mi

Hope this Helps!!!

3 0

2 years ago

The amount of pressure required to move a 6800 lb force with a 6" d piston is ___ psi.

Katena32 [7]

The pressure needed in PSI = Pounds of force needed divided by the cylinder Area
The Cylinder rod Area is 21.19 sq inches
Thus, the pressure= 6800/21.19
= 320.91 PSI

7 0

1 year ago

In the sport of parasailing, a person is attached to a rope being pulled by a boat while hanging from a parachute-like sail. A r

scoray [572]

Answer:

W = 506.75 N

Explanation:

tension = 2300 N

Rider is towed at a constant speed means there no net force acting on the rider.

hence taking all the horizontal force and vertical force in consideration.

net horizontal force:

F cos 30° - T cos 19° = 0

F cos 30° = 2300 × cos 19°

F = 2511.12 N

net vertical force:

F sin 30° - T sin 19°- W = 0

W = F sin 30° - T sin 19°

W = 2511.12 sin 30° - 2300 sin 19°

W = 506.75 N

8 0

2 years ago

If the radius of the sun is 7.001×105 km, what is the average density of the sun in units of grams per cubic centimeter? The vol

xenn [34]

Answer:

Average density of Sun is 1.3927 $\frac{g}{cm}$ .

Given:

Radius of Sun = 7.001 × $10^{5}$ km = 7.001 × $10^{10}$ cm

Mass of Sun = 2 × $10^{30}$ kg = 2 × $10^{33}$ g

To find:

Average density of Sun = ?

Formula used:

Density of Sun = $\frac{Mass of Sun}{Volume of Sun}$

Solution:

Density of Sun is given by,

Density of Sun = $\frac{Mass of Sun}{Volume of Sun}$

Volume of Sun = $\frac{4}{3} \pi r^{3}$

Volume of Sun = $\frac{4}{3} \times 3.14 \times [7.001 \times 10^{10}]^{3}$

Volume of Sun = 1.436 × $10^{33}$ $cm^{3}$

Density of Sun = $\frac{ 2\times 10^{33} }{1.436 \times 10^{33} }$

Density of Sun = 1.3927 $\frac{g}{cm}$

Thus, Average density of Sun is 1.3927 $\frac{g}{cm}$ .

4 0

1 year ago

A metallic sphere of radius 2.0 cm is charged with +5.0-μC+5.0-μC charge, which spreads on the surface of the sphere uniformly.

sladkih [1.3K]

Answer:

Explanation:

Potential due to a charged metallic sphere having charge Q and radius r on its surface will be

v = k Q / r . On the surface and inside the metallic sphere , potential is the same . Outside the sphere , at a distance R from the centre potential is

v = k Q / R

a ) On the surface of the shell , potential due to positive charge is

V₁ = $\frac{9\times10^9\times5\times10^{-6}}{6\times10^{-2}}$

On the surface of the shell , potential due to negative charge is

V₁ = $\frac{- 9\times10^9\times5\times10^{-6}}{6\times10^{-2}}$

Total potential will be zero . they will cancel each other.

b ) On the surface of the sphere potential

= $\frac{9\times10^9\times5\times10^{-6}}{2\times10^{-2}}$

= 22.5 x 10⁵ V

On the surface of the sphere potential due to outer shell

= $\frac{9\times10^9\times5\times10^{-6}}{5\times10^{-2}}$

= -9 x 10⁵

Total potential

=( 22.5 - 9 ) x 10⁵

= 13.5 x 10⁵ V

c ) In the space between the two , potential will depend upon the distance of the point from the common centre .

d ) Inside the sphere , potential will be same as that on the surface that is

13.5 x 10⁵ V.

e ) Outside the shell , potential due to both positive and negative charge will cancel each other so it will be zero.

5 0

1 year ago