A truck travels down the highway at a speed of 110 km/hr. How long does the trip last if the truck covered 2200km?

: In heavy rushIn heavy rush-hour traffic you drive in a straight line at 12 m/s for 1.5 minutes, then you have to stop for 3.5

zhuklara [117]

Guessing you want the average speed. We can multiple each speed by the time we spent going that speed, and them all together and then divide by the total time we spent in traffic to get the average speed. We spent a total of 7.5 minutes in traffic, so average speed = (12*1.5+0*3.5+15*2.5)/7.5 = 7.4 m/s

8 0

2 years ago

How are humans potentially vulnerable to some of these same chemicals? How are humans potentially vulnerable to some of these sa

Svetradugi [14.3K]

The 5 major unit processes include chemical coagulation, flocculation, sedimentation, filtration, and disinfection (described below). There are chemicals added to the water as it enters the various treatment processes.

6 0

2 years ago

A particle with charge 7.76×10^(−8)C is moving in a region where there is a uniform 0.700 T magnetic field in the +x-direction.

kodGreya [7K]

Answer:

The z-component of the force is $\= F_z = 0.00141 \ N$

Explanation:

From the question we are told that

The charge on the particle is $q = 7.76 *0^{-8} \ C$

The magnitude of the magnetic field is $B = 0.700\r i \ T$

The velocity of the particle toward the x-direction is $v_x = -1.68*10^{4}\r i \ m/s$

The velocity of the particle toward the y-direction is

$v_y = -2.61*10^{4}\ \r j \ m/s$

The velocity of the particle toward the z-direction is

$v_y = -5.85*10^{4}\ \r k \ m/s$

Generally the force on this particle is mathematically represented as

$\= F = q (\= v X \= B )$

So we have

$\= F = q ( v_x \r i + v_y \r j + v_z \r k ) \ \ X \ ( \= B i)$

$\= F = q (v_y B(-\r k) + v_z B\r j)$

substituting values

$\= F = (7.7 *10^{-8})([ (-2.61*10^{4}) (0.700)](-\r z) + [(5.58*10^{4}) (0.700)]\r y)$

$\= F= 0.00303\ \r j +0.00141\ \r k$

So the z-component of the force is $\= F_z = 0.00141 \ N$

Note : The cross-multiplication template of unit vectors is shown on the first uploaded image ( From Wikibooks ).

7 0

3 years ago

Two astronauts of mass 100 kg are 2 m apart in outer space. What is the

fredd [130]

The force of gravity between the astronauts is $1.67\cdot 10^{-7}N$

Explanation:

The magnitude of the gravitational force between two objects is given by:

$F=G\frac{m_1 m_2}{r^2}$

where :

$G=6.67\cdot 10^{-11} m^3 kg^{-1}s^{-2}$ is the gravitational constant

$m_1, m_2$ are the masses of the two objects

r is the separation between them

In this problem, we have two astronauts, whose masses are:

$m_1 = 100 kg\\m_2 = 100 kg$

While the separation between the astronauts is

r = 2 m

Substituting into the equation, we can find the gravitational force between the two astronauts:

$F=\frac{(6.67\cdot 10^{-11})(100)(100)}{2^2}=1.67\cdot 10^{-7}N$

Learn more about gravitational force:

brainly.com/question/1724648

brainly.com/question/12785992

#LearnwithBrainly

4 0

3 years ago

Read 2 more answers

Light does not move infinitely fast but has a finite speed. We normally use ""c"" to indicate the speed of light in science. Wri

Ede4ka [16]

Answer:

6.71 × 10^8 mi/hr

Explanation:

Light is usually defined as an electromagnetic wave that is comprised of a definite wavelength. It is of both types, visible and invisible. The light emitted from a source usually travels at a speed of about 3 × 10^8 meter/sec. This speed of light is commonly represented by the letter 'C'.

To write it in the metric system, it has to be converted into miles/hour.

We know that,

1 minute = 60 seconds

60 minutes = 1 hour

1 kilometer = 1000 meter

1 miles = 1.6 kilometer

Now,

= $\frac{3 \times\ 10^8 meter \times\ 60 sec \times\ 60 min}{1 sec \times\ 1 min \times\ 1 hr}$

= 1.08 × 10^12 m/ hr (meter/hour)

= $\frac{1.08 \times\ 10^{12} meter \times\ 1 km \times\ 1 miles}{1 hr \times\ 1000 meter \times\ 1.6 km}$

= 6.71 × 10^8 mi/hr (miles/hour)

Thus, the value for speed of light (C) in metric unit is 6.71 × 10^8 mi/hr.

5 0

3 years ago