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olga55 [171]
3 years ago
10

Carlos runs with velocity \vec{v}v →= (5.6 m/s, 29o north of east) for 10 minutes. How far to the north of his starting position

does Carlos end up? (Give your answer to the nearest meter)
Physics
1 answer:
romanna [79]3 years ago
7 0

Answer:

1629 metres

Explanation:

Given that Carlos runs with velocity v = (5.6 m/s, 29o north of east) for 10 minutes.

To the north, the velocity will be:

V = 5.6 × sin 29

V = 2.715 m/s

Convert the time to second

Time = 10 × 60 = 600

Using the formula below

Velocity = distance/ time

Substitute all the parameters into the formula

2.715 = distance/600

Distance = 2.715 × 600

distance = 1629 m

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A city of punjab has 15 percemt chance of wet weather on any given day. What is probability that it will take a week for it thre
sesenic [268]

Answer:

The probability that it will take a week for it three wet weather on 3 separate days is 0.06166 and its standard deviation is 0.9447

Explanation:

We are given that A city of Punjab has 15 percent chance of wet weather on any given day.

So, Probability of wet weather = 0.15

Probability of not being a wet weather = 1-0.15 =0.85

We are supposed to find probability that it will take a week for it three wet weather on 3 separate days

Total number of days in a week = 7

We will use binomial over here

n = 7

p =probability of failure = 0.15

q = probability of success=0.85

r=3

Formula :P(r=3)=^nC_r p^r q ^{n-r}

P(r=3)=^{7}C_{3} (0.15)^3 (0.85)^{7-3}\\P(r=3)=\frac{7!}{3!(7-3)!} (0.15)^3 (0.85)^{7-3}\\P(r=3)=0.06166

Standard deviation =\sqrt{n \times p \times q}

Standard deviation =\sqrt{7 \times 0.15 \times 0.85}

Standard deviation =0.9447

Hence The probability that it will take a week for it three wet weather on 3 separate days is 0.06166 and its standard deviation is 0.9447

4 0
3 years ago
If it takes 15 minutes to dry your hair with a 1.200 kW hair drier, how much energy is used in drying your hair?
Elodia [21]
1.2 kW * 0.25 h = 0.300 kWh
6 0
3 years ago
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Thermal energy depends on an object’s
oee [108]

The correct answer is D. I alread took this test.

4 0
3 years ago
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A 0.59 kg bullfrog is sitting at rest on a level log. how large is the normal force of the log on the bullfrog?
ladessa [460]
<span>The bullfrog is sitting at rest on the log. The force of gravity pulls down on the bullfrog. We can find the weight of the bullfrog due to the force of gravity. weight = mg = (0.59 kg) x (9.80 m/s^2) weight = 5.782 N The bullfrog is pressing down on the log with a force of 5.782 newtons. Newton's third law tells us that the log must be pushing up on the bullfrog with a force of the same magnitude. Therefore, the normal force of the log on the bullfrog is 5.782 N</span>
7 0
3 years ago
The sound intensity of a certain type of food processor in normally distributed with standard deviation of 2.9 decibels. If the
Maru [420]

Answer: (48.41,\ 52.19)

Explanation:

The confidence interval for population mean is given by :-

\overline{x}\pm z_{\alpha/2}\dfrac{\sigma}{\sqrt{n}}

Given : Sample size : n=9

Sample mean : \ovreline{x}=50.3\text{ decibels}

Standard deviation : \sigma=2.9\text{ decibels }

Significance level : \alpha=1-0.95=0.05

Critical value : z_{\alpha/2}=z_{0.025}=1.96

Now, the 95% confidence interval estimate of the (true, unknown) mean sound intensity of all food processors of this type :-

50.3\pm (1.96)\dfrac{2.9}{\sqrt{9}}\\\\\approx50.3\pm1.89\\\\=(50.3-1.89,\ 50.3+1.89)=(48.41,\ 52.19)

6 0
3 years ago
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