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lbvjy [14]
3 years ago
6

1. Which of the following is closest to having a mass of 1 kilogram?

Physics
1 answer:
Bingel [31]3 years ago
8 0

Answer:

C. One liter of water

Explanation:

One liter of water is closest to 1 kilogram. The other options don't make sense because a dime cannot be close to a mass of 1 kilogram, neither an average size banana or an average person

<em>PLEASE DO</em><em> </em><em>MARK ME</em><em> </em><em>AS BRAINLIEST</em><em> </em><em>IF MY</em><em> </em><em>ANSWER IS</em><em> </em><em>HELPFUL</em><em> </em><em>;</em><em>)</em><em> </em>

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What is the net force (including direction and magnitude) acting on the object on the above left?
Ber [7]

Answer:

Explanation:

the net force on the right left is 25 N and is directed upward

the net force on the left one is zero because 200N force act upward and 200N force act downward so both cancel each other and net force is zero

i hope this will help you

7 0
2 years ago
A psychologist claiming that a client's personal experiences and viewpoint influence behavior more than events in reality is usi
vodka [1.7K]
A psychologist who would claim that a client's personal experience and viewpoint influence behavior more than events in reality would probably use cognitive psychology mixed with developmental aspects to explain the behavior and personality of a person. 
5 0
2 years ago
Read 2 more answers
5. Find the mass of a car that is traveling at a velocity of 35 m/s West.
sattari [20]

Answer:

m = 9795.9 kg

Explanation:

v = 35 m/s

KE = 6,000,000 J

Plug those values into the following equation:

KE = \frac{1}{2} mv^{2}

6,000,000 J = (1/2)(35^2)m

---> m = 9795.9 kg

3 0
3 years ago
A car accelerates from rest at 1.0 m/s2 for 20.0 second along a straight road. It then moves at a constant speed for half an hou
Whitepunk [10]

Total distance = 36500 m

The average velocity = 19.73 m/s

<h3>Further explanation</h3>

Given

vo=initial velocity=0(from rest)

a=acceleration= 1 m/s²

t₁ = 20 s

t₂ = 0.5 hr = 1800 s

t₃= 30 s

Required

Total distance

Solution

State 1 : acceleration

\tt d=vo.t+\dfrac{1}{2}at^2\\\\d=\dfrac{1}{2}\times 1\times 20^2\rightarrow vo=0\\\\d=200~m

\tt vt=vo+at\\\\vt=at\rightarrow vo=0\\\\vt=1\times 20\\\\vt=20~m/s

State 2 : constant speed

\tt d=v\times t\\\\d=20\times 1800\\\\d=36000~m

State 3 : deceleration

\tt vt=vo+at\rightarrow vt=0(stop)\\\\vo=-at\\\\20=-a.30~s\\\\a=-\dfrac{2}{3}m/s^2(negative=deceleration)

\tt d=vot+\dfrac{1}{2}at^2\\\\d=20.30-\dfrac{1}{2}.\dfrac{2}{3}.30^2\\\\d=300~m

Total distance : state 1+ state 2+state 3

\tt 200 + 36000 + 300=36500~m

the average velocity = total distance : total time

\tt avg~velocity=\dfrac{36500}{20~s+1800~s+30~s}=19.73~m/s

4 0
2 years ago
11. A cyclist accelerates from 0 m/s to 10 m/s in 3 seconds. What is his acceleration ? Is this acceleration higher than that of
Marat540 [252]

a =  \frac{v - u}{t}

v = final velocity

u = initial velocity

t = time taken

the acceleration of the cyclist is

\frac{10 - 0}{3}  = 3.333333....

approximately 3.33 m/s^2

the acceleration of the car is

\frac{40 - 0 }{8}  = 5.0

5.0 m/s^2

5.0 > 3.33 \\ so \:  the \: answer  \: is \: no

6 0
3 years ago
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