What is the average velocity of the car from t=5 seconds to t=12 seconds?

Milk with a density of 1020 kg/m3 is transported on a level road in a 9-m-long, 3-m-diameter cylindrical tanker. The tanker is c

jeka57 [31]

Solution :

Given data is :

Density of the milk in the tank, $\rho = 1020 \ kg/m^3$

Length of the tank, x = 9 m

Height of the tank, z = 3 m

Acceleration of the tank, $a_x = 2.5 \ m/s^2$

Therefore, the pressure difference between the two points is given by :

$P_2-P_1 = -\rho a_x x - \rho(g+a)z$

Since the tank is completely filled with milk, the vertical acceleration is $a_z = 0$

$P_2-P_1 = -\rho a_x x- \rho g z$

Therefore substituting, we get

$P_2-P_1=-(1020 \times 2.5 \times 7) - (1020 \times 9.81 \times 3)$

$=-17850 - 30018.6$

$=-47868.6 \ Pa$

$=-47.868 \ kPa$

Therefore the maximum pressure difference in the tank is Δp = 47.87 kPa and is located at the bottom of the tank.

4 0

3 years ago

A car drives 20 miles north and then 3 miles south. what is the displacement or the car

kobusy [5.1K]

An:
Displacement is 16.

5 0

3 years ago

Read 2 more answers

In one of the classic nuclear physics experiments at the beginning of the 20th century, an alpha particle was accelerated toward

Vladimir79 [104]

Answer:

The answer is " $1.01 \times 10^{-13}$ "

Explanation:

Using the law of conservation for energy. Equating the kinetic energy to the potential energy.

$KE=U=\frac{kqq'}{r}\\\\$

Calculating the closest distance:

$\to r=\frac{kqq'}{KE}\\\\$

$=\frac{k(2e)(79e)}{KE}\\\\=\frac{k(2)(79)e^2}{KE}\\\\=\frac{9.0\times 10^9 \ N \cdot \frac{m^2}{c}(2)(79)(1.6 \times10^{-19} \ C)^2}{(2.25\ meV) (\frac{1.6 \times 10^{-13} \ J}{1 \ MeV})}\\\\$

$=\frac{9.0\times 10^9 \times 2\times 79\times 1.6 \times10^{-19}\times 1.6 \times10^{-19} }{(2.25 \times 1.6 \times 10^{-13}) }\\\\=\frac{3,640.32\times 10^{-29}}{3.6 \times 10^{-13} }\\\\=\frac{3,640.32}{3.6} \times 10^{-16}\\\\=1011.2 \times 10^{-16}\\\\=1.01 \times 10^{-13}$

5 0

3 years ago

The main purpose of an air bag is to stop a passenger during a car accident in a greater amount of time than if the air bag were

Simora [160]

Answer:

a) 45571 N

b) 22786 N

c) 4557 N

Explanation:

Since the goal of the airbag is helping the person to stop after the collision in a greater time, this means that the change in momentum must finish when this is just zero.
In other words, the change in momentum, must be equal to the initial one, but with opposite sign.

$\Delta p = - p_{o} = -m*v = -55 kg*29m/s = -1595 kgm/s (1)$

Now, just applying the original form of Newton's 2nd Law, we know that this change in momentum must be equal to the impulse needed to stop the person:

$\Delta p = F* \Delta t (2)$

So, as we know the magnitude of Δp from (1) and we have different Δt as givens, we can get the different values of F (in magnitude) required to stop the person for each one of them, as follows:

$F_{1} = \frac{\Delta p}{\Delta t_{1}} = \frac{1595kgm/s}{0.035s} = 45571 N (3)$