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borishaifa [10]
3 years ago
11

A car with an initial speed of 6.5 m/s accelerates at a uniform rate of 0.92 m/s2 for 3.6 s. Find the final speed of the car dur

ing this time.
Physics
1 answer:
geniusboy [140]3 years ago
8 0
U = 6.5 m/s, initial speed
t = 3.6 s, time
a = 0.92 m/s², acceleration

Let v = the final velocity.
Then
v = u +at
v = (6.5 m/s) + (0.92 m/s²)*(3.6 s) = 9.812 m/s

Answer: 9.81 m/s 
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What is the idea behind the law of multiple proportions?
Akimi4 [234]

Answer:

The law of multiple proportions states that when two elements can combine in different ratios to form different compounds, the masses of the element combining with the fixed mass of another element result in whole number ratios. This shows that the law of multiple proportions is followed

Explanation:

5 0
4 years ago
A 70- kg bicycle rides his 9.8- kg bicycle with a speed of 16 m/ s. What is the magnitude of the braking force of the bicycle co
Rus_ich [418]

Answer:

F = -319.2 N

Explanation:

Given that,

The mass of a bicyclist, m = 70 kg

Mass of the bicycle = 9.8 kg

The speed of a bicycle, v = 16 m/s

We need to find the magnitude of the braking force of the bicycle come to rest in 4.0 m.

The braking force is given by :

F=ma\\\\=\dfrac{m(v-u)}{t}\\\\=\dfrac{(70+9.8)(0-16)}{4}\\\\=-319.2\ N

So, the required force is 319.2 N.

3 0
3 years ago
A 0.683 kg mass moves in SHM at the end of a spring. It takes 1.41 s to move from the position with the spring fully extended to
dsp73

Answer:

Spring constant of the spring will be equal to 9.255 N /m          

Explanation:

We have given mass m = 0.683 kg

Time taken to complete one oscillation is given T = 1.41 sec

We have to find the spring constant of the spring

From spring mass system time period is equal to T=2\pi \sqrt{\frac{m}{K}}, here m is mass and K is spring constant

So 1.41=2\times 3.14\sqrt{\frac{0.683}{K}}

0.2245=\sqrt{\frac{0.683}{K}}

Squaring both side

0.0504=\frac{0.4664}{K}

K=9.255N/m

So spring constant of the spring will be equal to 9.255 N /m

7 0
3 years ago
A child swings on a playground swing. How many times does the child swing through the swing's equilibrium position during the co
Karolina [17]
<h2>The child swing through the swing's equilibrium position 6 times during the course of 3 periods.</h2>

Explanation:

One period means time taken to complete one revolution.

In case of swings in one period time it travels the same position through two times.

Here we need to find how many times does the child swing through the swing's equilibrium position during the course of 3 period(s) of motion.

For 1 period = 2 times

For 3 periods = 3 x For 1 period

For 3 periods = 3 x 2 times

For 3 periods = 6 times

The child swing through the swing's equilibrium position 6 times during the course of 3 periods.

3 0
3 years ago
An electric motor consumes 9.00 kj of electrical energy in 1.00 min. if one-third of this energy goes into heat and other forms
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