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Westkost [7]
3 years ago
13

a driver travels 135 km, east in 1.5 h, stops for 45 minutes for lunch, and then resumes driving for the next 2.0 h through a di

splacement of 215 km, east. what is the driver’s average velocity?
Physics
1 answer:
daser333 [38]3 years ago
7 0

Answer:

v = 98.75 km/h

Explanation:

Given,

The distance driver travels towards the east, d₁ = 135 km

The time period of the travel, t₁ = 1.5 h

The halting time, tₓ = 46 minutes

The distance driver travels towards the east, d₂ = 215 km

The time period of the travel, t₁ = 2 h

The average speed of the vehicle before stopping

                                    v₁ = d₁/t₁

                                        = 135/1.5

                                       = 90 km/h

The average speed of vehicle after stopping

                                    v₂ = d₂/t₂

                                         = 215/2

                                        = 107.5 km/h

The total average velocity of the driver

                                      v = (v₁ +v₂) /2

                                         = (90 + 107.5)/2

                                         = 98.75 km/h

Hence, the average velocity of the driver, v = 98.75 km/h

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3 years ago
Which example identifies a change in motion that produces acceleration?
kifflom [539]
Acceleration is the rate at which an object changes its velocity. It defines how much the velocity is changing. The acceleration can be negative and positive. Negative acceleration is when the object slows down, while positive while the object goes faster.
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6 0
3 years ago
Read 2 more answers
A uniform, spherical, 1900.0 kg shell has a radius of 5.00 m. Find the gravitational force this shell exerts on a 1.80 kg point
Mandarinka [93]

Answer:

F=9.09\times 10^{-9} N

Explanation:

We are given that

Mass of spherical shell,m_1=1900 kg

Mass=m_2=1.80 kg

Radius of shell=r=5 m

Distance between two masses=r=5.01 m

Because distance measure from center .

Gravitational force

F=G\frac{m_1m_2}{r^2}

G=6.67\times 10^{-11} Nm^2/kg^2

Using the formula

F=6.67\times 10^{-11}\times \frac{1900\times 1.80}{(5.01)^2}

F=9.09\times 10^{-9} N

Hence,the gravitational force =F=9.09\times 10^{-9} N

6 0
3 years ago
An object travels 7.5 m/s toward the west . Under the influence of a constant net force of 5.2 kN, it comes to rest in 3.2 s. Wh
Softa [21]

Answer:

m = 2218.67 kg

Explanation:

It is given that,

Initial velocity, u = 7.5 m/s

Final speed of an object, v = 0 (at rest)

Force, F = 5.2 kN

Time, t = 3.2 s

We need to find the mass of the object. Force acting on an object is given by :

F = ma

m is mass, a is acceleration

F=\dfrac{m(v-u)}{t}\\\\m=\dfrac{Ft}{v-u}\\\\m=\dfrac{5.2\times 10^3\times 3.2}{0-7.5}\\\\m=2218.67\ kg

So, the mass of the object is 2218.67 kg

5 0
3 years ago
A rocket with a mass of 62,000 kg (including fuel) is burning fuel at the rate of 150 kg/s and the speed of the exhaust gases is
mezya [45]

Answer:

h≅ 58 m

Explanation:

GIVEN:

mass of rocket M= 62,000 kg

fuel consumption rate =  150 kg/s

velocity of exhaust gases v= 6000 m/s

Now thrust = rate of fuel consumption×velocity of exhaust gases

=6000 × 150 = 900000 N

now to need calculate time t = amount of fuel consumed÷ rate

= 744/150= 4.96 sec

applying newton's law

M×a= thrust - Mg

62000 a=900000- 62000×9.8

acceleration a= 4.71 m/s^2

its height after 744 kg of its total fuel load has been consumed

h= \frac{1}{2}at^2

h= \frac{1}{2}4.71\times4.96^2

h= 58.012 m

h≅ 58 m

4 0
3 years ago
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