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3 years ago

If a car travels 30 mi. north for 30 min., 60 mi. east for 1.0 hour, and 30 mi. south for 30 min., what is the average speed

Physics

1 answer:

Ainat [17]3 years ago

3 0

Answer: 60mph

Explanation:

Given the following :

First leg travel:

Distance = 30 miles

Time of travel= 30 minutes = 0.5 hour

Second leg travel:

Distance = 60 miles

Time of travel = one hour

Average speed :

Speed = total Distance / time of travel

Total distance in miles = (30 + 60) miles = 90 miles

Total time of travel = 1 hour + 0.5 hour = 1.5 hours

Average speed = total distance traveled / total travel time

Average speed = 90 miles / 1.5 hours

Average speed = 60 miles / hour

= 60mph

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Calculate the height of a column of carbon tetrachloride, CCl4(l), with a density of 1.59 g/mL that exerts the same pressure as

julia-pushkina [17]

Answer:

1.30 m

Explanation:

The pressure (P) exerted by a column of a liquid can be calculated with the following expression:

$P=\rho \times g \times h$

where,

ρ is the density of the liquid

g is the gravity

h is the height of the column

If both liquids exert the same pressure:

$PHg = PCCl_{4}\\\rho Hg \times g \times hHg = \rho CCl_{4} \times g \times hCCl_{4}\\\rho Hg \times hHg = \rho CCl_{4} \times hCCl_{4}\\hCCl_{4 = \frac{\rho Hg \times hHg}{\rho CCl_{4}} =\frac{13.6g/mL \times 15.2cm}{1.59g/mL} = 130cm=1.30m$

6 0

3 years ago

Can someone please help me!

Dimas [21]

Hello there!

The correct answer is 7.24

See picture below for the steps!

As always, I am here to help!

4 0

3 years ago

A ball is thrown straight up at 20 m. What is the balls velocity as it hits the ground?

deff fn [24]

Answer:

The velocity of the ball as it hit the ground = 19.799 m/s

Explanation:

Velocity: Velocity of a body can be defined as the rate of change of displacement of the body. The S.I unit of velocity is m/s. velocity is expressed in one of newtons equation of motion, and is given below.

v² = u² + 2gs.......................... Equation 1

Where v = the final velocity of the ball, g = acceleration due to gravity, s = the height of the ball

Given: s = 20 m, u = 0 m/s

Constant: g = 9.8 m/s²

Substituting these values into equation 1,

v² = 0 + 2×9.8×20

v² = 392

v = √392

v = 19.799 m/s.

Therefore the velocity of the ball as it hit the ground = 19.799 m/s

5 0

3 years ago

Particles 1 and 2 of charge q1 = q2 = +3.20 × 10−19 C are on a y axis at distance d = 17.0 cm from the origin. Particle 3 of cha

mel-nik [20]

Answer:

(a) 0.17 m

(b) 5.003 m

(d) 7.37 × $10^{-29}$ N

Explanation:

(a) The minimum value of $x$ will occur when q3 = 0 m or at origin and q1, q2 are at 0.17 m so the distance between q3 and q1, q2 is 0.17 m, therefore the minimum value of x= 0.17 m.

(b) The maximum value of x will occur when q3 = 5 m because it is said in the question that 5 is the maximum distance travelled by q3. To find the hypotenuse i.e. the distance between q3 and q1,q2, we use Pythagoras theorem.

$h^{2} = b^{2} + p^{2}$

$h^{2} = 5^{2} + 0.17^{2} \\h = \sqrt{} 25.03\\h= 5.002 m$

Hence, the maximum distance is 5.002 m

Minimum Distance = 0.17 m

For electrostatic force= $F=\frac{kq1q2}{x^{2} }$

$F=\frac{9 x 10^{9} x3.2x10^{-19}x 6.4x10^{-19} }{0.17^{2} }$

$F= 6.38$ × $10^{-26} N$

(d) For maximum magnitude, we use the maximum distance calculated in (b)

Maximum Distance = 5.002 m

Using the formula for electrostatic force again:

F = $\frac{9x10^{9}x3.2x10^{-19}x6.4x10^{-19} }{5.002^{2} } }$

F= 7.37× $10^{-29$ N

4 0

2 years ago

A tennis player tosses a tennis ball straight up and then catches it after 2.07 s at the same height as the point of release.

prisoha [69]

(a) The ball will accelerate at a speed of 9.81 m/s² while it is in flight.

(b) The ball will be moving at a speed of 0 m/sec when it reaches its highest point.

(d) It will reach a maximum height of 21.0 m.

<h3 /><h3>What is speed?</h3>

Speed is defined as the change in distance with regard to time. A scalar quantity, speed. It is a temporal element, m/sec is its unit.

The information provided in the issue is;

u is the fall's beginning speed in m/sec;

h is the fall's distance

g is the fall's acceleration or 9.81 m/sec²

a. The ball's speed while in flight is equal to the acceleration caused by gravity, which is 9.81 m/s².

b. When the ball reaches its highest point, its velocity will be zero.

c. The ball's starting velocity is;

v=u-gt

0=u-gt

u=gt

u=9.81 m/s² × 2.07 sec

u=20.30 m/s

d. The formula is used to determine the highest height it can reach.

$\rm u^2 = - 2gh \\\\ (20.30)^2 = -2 \times (-9.8) \times h \\\\ 412.09 = 19.6 \times h\\\\ h = 412.09/19.6\\\\ h = 21.00 m$

As a result, the ball's greatest height, beginning velocity, and acceleration are all 9.81 meters per second, zero meters per second, 20.30 meters per second, and 21 meters accordingly.

To learn more about the velocity, refer to the link: brainly.com/question/862972.

#SPJ1

3 0

2 years ago