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Rudiy27
2 years ago
10

WER Câu 11. Ôtô bắt đầu chuyển động từ trạng thái đứng yên và đạt đến tốc độ c 70 ft/s) sau khi đi được quãng đường d – 400 ft)

dọc theo đường thẳng. Xác định gia tốc không đổi của ôtô và thời gian để ô tô đạt được tốc độ trên​
Physics
1 answer:
Andreyy892 years ago
6 0

Answer:

sorry I don't understand u writing

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What is the kinetic energy k of an electron with momentum 1.05×10−24 kilogram meters per second?
fiasKO [112]
Momentum = mv
where m is the mass of an electron and v is the velocity of the electron.

v = momentum ÷ m
   = (1.05×10∧-24)÷(9.1×10∧-31) = 1,153,846.154 m/s

kinetic energy = (mv∧2)÷2
                       = (9.1×10∧-31 × 1,153,846.154∧2) ÷2
                      = (1.21154×10∧-18) ÷ 2
                      = 6.05769×10∧-19 J
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3 years ago
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3. Discuss the main purpose of Experiment" Explain the greatest advantage of experiment in psychology?
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Answer:

Social psychologists utilize experimental techniques to study how people are influenced by groups. Health psychologists rely on experimentation and research to better understand the factors that contribute to wellness and disease.

Explanation:

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3 years ago
How does geomagnetism help scientist understand the motion of earth’s plates?
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Answer:

As the "plates" on each side of ridges in the seafloor are pulled away, lava comes up from the middle, hardens and "records" the current magnetic field.

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3 years ago
An object weighing 49 N is dropped from a height of 30 m. It is found to be moving with a velocity of 24m/s just before it hits
faltersainse [42]

The  frictional force will be 0.22N.

<h3>What is Frictional force?</h3>

Frictional force is the force generated between two surfaces that are in contact and slide against each other.

Given,

Weight=4N

mass =4.9/9.8=0.5kg

Hieght =30m

velocity=24m/s

Acceration , v²-u²=2as

24²/2×30 =a , u is zero

 a= 1.5m/s²



By Using conservation  of energy ,

30F+1/2mv²=mgh

30F=1150-144

F= 6/30

F=0.2N

The force will be 0.2N

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6 0
2 years ago
A 50-kg satellite circles the Earth in an orbit with a period of 120 min. What minimum energy is required to change the orbit to
uysha [10]

Answer: 2.94×10^8 J

Explanation:

Using the relation

T^2 = (4π^2/GMe) r^3

Where v= velocity

r = radius

T = period

Me = mass of earth= 6×10^24

G = gravitational constant= 6.67×10^-11

4π^2/GMe = 4π^2 / [(6.67x10^-11 x6.0x10^24)]

= 0.9865 x 10^-13

Therefore,

T^2 = (0.9865 × 10^-13) × r^3

r^3 = 1/(0.9865 × 10^-13) ×T^2

r^3 = (1.014 x 10^13) × T^2

To find r1 and r2

T1 = 120min = 120*60 = 7200s

T2 = 180min = 180*60= 10800s

Therefore,

r1 = [(1.014 x 10^13)7200^2]^(1/3) = 8.07 x 10^6 m

r2 = [(1.014 x 10^13)10800^2]^(1/3) = 10.57 x 10^6 m

Required Mechanical energy

= - GMem/2 [1/r2 - 1/r1]

= (6.67 x 10^-11 x 6.0 x 10^24 * 50)/2 * [(1/8.07 × 10^-6 )- (1/10.57 × 10^-6)]

= (2001 x 10^7)/2 * (0.1239 - 0.0945)

= (1000.5 × 10^7) × 0.0294

= 29.4147 × 10^7 J

= 2.94 x 10^8 J.

6 0
2 years ago
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