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Andreyy89
2 years ago
7

A person travels by car from one city to an-

Physics
1 answer:
Mademuasel [1]2 years ago
5 0
<h2>Answer:</h2>

<h2>Explanation:</h2>

First, let's refer to the distance formula:

d=v*t, where d is distance, v is velocity or speed and t is time.

Now, let's find the distance covered by each individual speed that the car had:

<h3>1. Speed 1.</h3>

In order to use the formula, we need to convert minutes into hours since the speed is given in km/h.

21.1 min/60= 0.35 h.

Now, apply the distance formula.

d=(0.35h)*(86.8km/h)= 30.38 km.

<h3>2. Speed 2.</h3>

Convert minutes to hours again and do the same calculations.

10.6min/60=0.18h

d=(0.18h)*(106km/h)= 19.08 km.

<h3>3. Speed 3.</h3>

36.5min/60= 0.61h

d=(0.61h)*(30.9km/h)= 18.85 km.

<h3>4. Obtain the total distance.</h3>

The total distance must be given by the addition of all individual distances traveled by the car on each speed:

Total distance= 30.38 km + 19.08 km + 18.85 km= 68.31 km.

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The three stages of a train route took 1 hour ,2 hours ,and 4 hours . The first two stages were 80km and 200km of the train aver
luda_lava [24]

Answer:

the third stage was 480 km long

Explanation:

Stage 1:

Time = 1 hours

Speed = 80km

Stage 2:

Time =  2 hours

Speed = 200km

Stage 3:

Time =  4 hours

Let the Distance at the stage 3 be x

Average speed of the train route = 100 km/h

So

\frac{ \text{speed at stage 1} + \text{speed at stage 2} + \text{speed at stage 3}}{3} = 0

\frac{ \text{speed at stage 1} + \text{speed at stage 2} + \text{speed at stage 3}}{3} = 100

Lets find the speed at stage 1

Speed =  \frac{Distance }{Time}

Speed =  \frac{80}{1}

Speed 1= 80 km/hr

The speed at stage 2

Speed =  \frac{Distance }{Time}

Speed =  \frac{200}{2}

Speed 2  = 100 km/hr

The speed at stage 3

Speed =  \frac{Distance }{Time}

Speed =  \frac{x}{4}

Speed 3  = \frac{x}{4}

we kow that average is ,

\frac{ \text{speed 1} + \text{speed 2} + \text{speed 3}}{3} = 100

\frac{ 80 + 100+ \frac{x}{4} }{3} = 100

\frac{ 180 + \frac{x}{4} }{3} = 100

\frac{ \frac{720 +x}{4} }{3} = 100

\frac{720 +x}{4} \times \frac{1}{3} = 100

\frac{720 +x}{12} = 100

720 +x = 100 \times 12

720 +x = 1200

x = 1200- 720

x = 480

6 0
3 years ago
A 12 kg object speeds up from an initial velocity of 10 m:s-1
amid [387]

Momentum = m • v

Original momentum = m • 10 m/s north

Final momentum = m • 15 m/s north

Change = m • (15 - 10) m/s north

Change = m • +5 m/s north

Change = +60 kg-m/s north

5 0
3 years ago
When applying a horizontal force of 30N, an object of mass 6 kg accelerates at 4m/s2. The force of friction on the surface must
Rus_ich [418]
Using Newton's Second Law, F = ma, where F is the net force

So the net force is:

F = (6kg)(4m/s^2) = 24N

Since you are applying a horizontal force of 30N, we can find the force of friction by the difference of the net force and the applied force.

30N-24N = 6N

F_{f} = 6N
4 0
3 years ago
All are examples of electric forces except _________. A. a neutron pushing on another neutron B. an electron pushing on a proton
Contact [7]

Answer:

A) a neutron pushing on another neutron.

Explanation:

Neutrons have no charge, therefore, there is no electric force among them. Protons and electrons on the other hand do have electric charge (electrons negative charge and protons positive charge) that generate electric forces between them that can be repelling forces (if the charges are of the same sign), or attractive forces (if the charges are of opposite signs).

6 0
3 years ago
During the execution of a play, a football player carries the ball for a distance of 33 m in the direction 58° north of east. To
podryga [215]

Answer:28 m

Explanation:

Given

Direction is 58^{\circ} North of east i.e. 58 ^{\circ} with x axis

Also ball moved by 33 m

therefore its east component is 33cos58=17.48 m

Northward component =33sin58=27.98 m\approx 28 m

8 0
3 years ago
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