Question 2 (1 point)

A 1369.4 kg car is traveling at 28.9 m/s when the driver takes his foot off the gas pedal. It takes 5.1 s for the car to slow do

Darya [45]

Answer:

F = 2389.603 N

Explanation:

Given:

Mass m = 1,369.4 kg

Initial velocity u = 28.9 m/s

Final velocity v = 20 m/s

Time t = 5.1 s

Find:

Net force

Computation:

a = (v - u)/t

a = (20 - 28.9)/5.1

a = -1.745 m/s²

F = ma

F = (1369.4)(1.745)

F = 2389.603 N

7 0

3 years ago

Three joggers are running along straight lines as follows: Jogger A, with a mass of 55.2kg, is traveling along the line y=6.00m

frutty [35]

Answer:

L = - 1361.591 k Kgm/s

Explanation:

Given

mA = 55.2 Kg

vA = 3.45 m/s

rA = 6.00 m

mB = 62.4 Kg

vB = 4.23 m/s

rB = 3.00 m

mC = 72.1 Kg

vC = 4.75 m/s

rC = - 5.00 m

then we apply the equation

L = (mv x r)

⇒ LA = mA*vA x rA = 55.2 *(3.45 i)x(6 j) = (1142.64 k) Kgm/s

⇒ LB = mB*vB x rB = 62.4 *(4.23 j)x(3 i) = (- 791.856 k) Kgm/s

⇒ LC = mC*vC x rC = 72.1 *(- 4.75 j)x(- 5 i) = (- 1712.375 k) Kgm/s

Finally, the total counterclockwise angular momentum of the three joggers about the origin is

L = LA + LB + LC = (1142.64 - 791.856 -1712.375) k Kgm/s

L = - 1361.591 k Kgm/s

7 0

3 years ago

Two thin 80.0-cm rods are oriented at right angles to each other. Each rod has one end at the origin of the coordinates, and one

kogti [31]

Answer:

The net force on the electron is given as:

F = 1.35 x 10⁻¹³ N j - 1.35 x 10⁻¹³ N i

Explanation:

Given:

charge on rod along x-axis = Q₁ = -15 x 10⁻⁶ C

charge on rod along y-axis = Q₂ = 15 x 10⁻⁶ C

distance of electron from rod 1 = r₁ = 0.4 m

distance of electron from rod 1 = r₂ = 0.4 m

charge on electron = q = -1.6 x 10⁻¹⁹ C

ε° = 8.85 x 10⁻¹² C²/Nm²

Electric force on charge due to rod 1:

F₁ = qE = 1/4πε°(qQ₁/r₁²)

F₁ = (9 x 10⁹ x -1.6 x 10⁻¹⁹ x -15 x 10⁻⁶)/0.4²

F₁ = 1.35 x 10⁻¹³ N

Negative negative repels each other so the rod will Force the electron in positive y-direction.

F₁ = 1.35 x 10⁻¹³ N j

Electric force on charge due to rod 2:

F₂ = qE = 1/4πε°(qQ₂/r₂²)

F₂ = (9 x 10⁹ x -1.6 x 10⁻¹⁹ x 15 x 10⁻⁶)/0.4²

F₂ = - 1.35 x 10⁻¹³ N

Opposite charges attract each other so the rod will force the electron in negative x-direction.

F₂ = - 1.35 x 10⁻¹³ N i

Net Force:

F = F₁ + F₂

F = 1.35 x 10⁻¹³ N j - 1.35 x 10⁻¹³ N i

4 0

3 years ago

What is the gauge pressure of the water right at the point p, where the needle meets the wider chamber of the syringe? neglect t

Helen [10]

Missing details: figure of the problem is attached.

We can solve the exercise by using Poiseuille's law. It says that, for a fluid in laminar flow inside a closed pipe,

$\Delta P = \frac{8 \mu L Q}{\pi r^4}$

where:

$\Delta P$ is the pressure difference between the two ends

$\mu$ is viscosity of the fluid

L is the length of the pipe

$Q=Av$ is the volumetric flow rate, with $A=\pi r^2$ being the section of the tube and $v$ the velocity of the fluid

r is the radius of the pipe.

We can apply this law to the needle, and then calculating the pressure difference between point P and the end of the needle. For our problem, we have:

$\mu=0.001 Pa/s$ is the dynamic water viscosity at $20^{\circ}$

L=4.0 cm=0.04 m

$Q=Av=\pi r^2 v= \pi (1 \cdot 10^{-3}m)^2 \cdot 10 m/s =3.14 \cdot 10^{-5} m^3/s$

and r=1 mm=0.001 m

Using these data in the formula, we get:

$\Delta P = 3200 Pa$

However, this is the pressure difference between point P and the end of the needle. But the end of the needle is at atmosphere pressure, and therefore the gauge pressure (which has zero-reference against atmosphere pressure) at point P is exactly 3200 Pa.

8 0

3 years ago

All of the following are categories of individual activities except?

aliina [53]

B. Hand-eye Coordination because it can be used in multiple activities

3 0

3 years ago