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Sergeu [11.5K]
2 years ago
11

You walk 12.0 m West and then 4.00 m south. What is your displacement?​

Physics
1 answer:
Karo-lina-s [1.5K]2 years ago
7 0

Answer:

12.6 (3 sig. fig.)

Explanation:

(12^2 + 4^2)^1/2 = 12.6 (3 sig. fig.)

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oksano4ka [1.4K]

the answer of this question is helium

4 0
3 years ago
A car accelerates from 300 km/h to 140 km/h in 2.53 seconds. what is the distance covered?
Snowcat [4.5K]

Answer:

Acceleration = Change in Velocity/Time

Change in Velocity = 36-18 = 18 km/h=5 m/s

Time= 5 Seconds

Acceleration = 5/5= 1 m/s2

Equation of motion,s=ut+(1/2)at2

u=18 km/h=5 m/s

t=5 s

a=1 m/s2

s= (5*5)+(1/2*1*5*5)

s=25+12.5 i.e., s=37.5 m

Hope you are clear with my explanations

7 0
3 years ago
The first step when doing<br> an<br> investigation is the observe a situation. True or false?
Sunny_sXe [5.5K]

Answer:

True! First step is to make objective observations.

6 0
3 years ago
Read 2 more answers
In mammals, the weight of the heart is approximately 0.5% of the total body weight. Write a linear model that gives the heart we
hammer [34]

Answer:

1201 lbs

Explanation:

Given that in mammals, the weight of the heart is approximately 0.5% of the total body weight.

Let the weight of the heart of a mammal be H

And the weight of the total body be B

The linear model that can gives the heart weight in terms of the total body weight will be:

H = 0.005B

B.) To find the weight of the heart of a whale whose weight is 2.402 × 105 lbs, substitute the whole weight in the formula.

H = 0.005 × 2.402 × 10^5

H = 1201 lbs

Therefore, the weight of the heart of the whale is 1201 lbs

8 0
3 years ago
a whistle you use to call your hunting dog has a frequency of 21 khz, but your dog is ignoring it. you suspect the whistle may n
laila [671]

To solve this problem we will apply the concepts related to the Doppler effect. According to this concept, it is understood as the increase or decrease of the frequency of a sound wave when the source that produces it and the person who captures it move away from each other or approach each other. Mathematically this can be described as

f = f_0 (\frac{v-v_0}{v})

Here,

f_0 = Original frequency

v_0 = Velocity of the observer

v = Velocity of the speed

Our values are,

v = 340m/s \rightarrow \text{Speed of sound}

f = 20kHz \rightarrow \text{Apparent frequency}

f_0 = 21kHz \rightarrow \text{Original frequency}

Using the previous equation,

f = f_0 (\frac{v-v_0}{v})

Rearrange to find the velocity of the observer

v_0 =v (1-\frac{f}{f_0})

Replacing we have that

v_0= (340m/s)(1-\frac{20kHz}{21kHz})

v_0 = 16.19m/s

Therefore the velocity of the observer is 16.2m/s

4 0
3 years ago
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