Calculate the force applied (in newtons) if a pressure of 2000Pa is acting on an area of 3m2.

2200 kg semi truck driving down the highway has lost control. The truck rolls across the median and into oncoming traffic. The t

serious [3.7K]

Answer:

The semi truck travels at an initial speed of 69.545 meters per second downwards.

Explanation:

In this exercise we see a case of an entirely inellastic collision between the semi truck and the car, which can be described by the following equation derived from Principle of Linear Momentum Conservation: (We assume that velocity oriented northwards is positive)

$m_{S}\cdot v_{S}+m_{C}\cdot v_{C} = (m_{S}+m_{C})\cdot v$ (1)

Where:

$m_{S}$ , $m_{C}$ - Masses of the semi truck and the car, measured in kilograms.

$v_{S}$ , $v_{C}$ - Initial velocities of the semi truck and the car, measured in meters per second.

$v$ - Final speed of the system after collision, measured in meters per second.

If we know that $m_{S} = 2200\,kg$ , $m_{C} = 2000\,kg$ , $v_{C} = 45\,\frac{m}{s}$ and $v = -15\,\frac{m}{s}$ , then the initial velocity of the semi truck is:

$m_{S}\cdot v_{S} = (m_{S}+m_{C})\cdot v -m_{C}\cdot v_{C}$

$v_{S} = \frac{(m_{S}+m_{C})\cdot v - m_{C}\cdot v_{C}}{m_{S}}$

$v_{S} = \left(1+\frac{m_{C}}{m_{S}} \right)\cdot v - \frac{m_{C}}{m_{S}} \cdot v_{C}$

$v_{S} = v +\frac{m_{C}}{m_{S}}\cdot (v-v_{C})$

$v_{S} = -15\,\frac{m}{s}+\left(\frac{2000\,kg}{2200\,kg} \right) \cdot \left(-15\,\frac{m}{s}-45\,\frac{m}{s} \right)$

$v_{S} = -69.545\,\frac{m}{s}$

The semi truck travels at an initial speed of 69.545 meters per second downwards.

3 0

2 years ago

A Thomson's gazelle can run at very high speeds, but its acceleration is relatively modest. A reasonable model for the sprint of

NikAS [45]

Answer:

The gazelles top speed is 27.3 m/s.

Explanation:

Given that,

Acceleration = 4.2 m/s²

Time = 6.5 s

Suppose we need to find the gazelles top speed

The speed is equal to the product of acceleration and time.

We need to calculate the gazelles top speed

Using formula of speed

$v=at$

Where, v = speed

a = acceleration

t = time

Put the value into the formula

$v=4.2\times6.5$

$v=27.3\ m/s$

Hence, The gazelles top speed is 27.3 m/s.

6 0

3 years ago

20% Part (a) Use an "E Field Sensor" and move it along either equipotential. What can you say about the E field along an equipot

Alex

Answer:

b. Constant magnitude, but varying direction, perpendicular to the equipotential.

Explanation:

As we know that the relation between electric field and electric potential is given as

$\Delta E = -\frac{dV}{dr}$

here if we say that potential is constant because electric field sensor is moving along equi-potential line.

Then we will say

V = constant

so we have

$\Delta E = 0$

so electric field will remain constant always in magnitude and always remains perpendicular to the surface

so we have

b. Constant magnitude, but varying direction, perpendicular to the equipotential.

6 0

3 years ago

The speed of light in a vacuum is approximately 2.99 × 108 m/s, and the speed of light through a piece of glass is approximately

exis [7]

Option (b) is correct.The index of refraction for the glass is 1.52

Explanation:

velocity of light in vacuum= C= 2.99 x 10⁸m/s

Velocity of light in glass = V= 1.97 x 10⁸m/s

The refractive index is given by n= $\frac{c}{v}$

n= 2.99 x 10⁸/1.97 x 10⁸m/s

n= 1.52

Thus the refractive index of glass is 1.52

6 0

3 years ago

Read 2 more answers

If, while standing on the bank of a stream, you wished to spear a fish swimming in the water out in front of you, would you aim

Serggg [28]

Answer:

a) below the observed position

b) directly at the observed position

Explanation:

If I'm standing on the bank of a stream, and I wish to spear a fish swimming in the water out in front of me, I would aim below the observed fish to make a direct hit. This is because the phenomenon of refraction of light in water causes the light coming from the fish is refract away from the normal as it passes into the air and into my eyes.

If I'm to zap the fish with a taser, I would aim directly at the observed fish because the laser (a form of concentrated light waves) will refract into the water, taking the same path the light from the fish took to get to my eyes.

3 0

3 years ago