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kipiarov [429]
2 years ago
8

Hai ô tô khởi hành cùng lúc từ A,B cách nhau 20 km chuyển động theo hướng từ A đến B vs vận tốc lần lượt là 60km/h và 40km/h

Physics
1 answer:
sdas [7]2 years ago
6 0

Answer:

áp dụng công thức v = s/t

s là dộ dài qduong

v là vận tốc

t là thời gian

xuyên suốt 2 câu hỏi đều dùng công thức này

Explanation:

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The density of a certain type of plastic is 0.75 g/cm3 . If a sheet of this plastic is 10.0 m long, 1.0 m wide, and 1 cm thick,
Oksanka [162]

Answer:

7.5 g

Explanation:

8 0
2 years ago
A technique in which the muscles are stretched by an outside force is called _____.
geniusboy [140]
A technique in which the muscles are stretched by an outside force is called Passive Stretching 
6 0
3 years ago
A truck traveling at a constant speed of 28 m/s passes a more slowly moving car. The instant the truck passes the car, the car b
choli [55]

Answer:

the velocity of car when it passes the truck is u = 16.33 m/s

Explanation:

given,

constant speed of truck  = 28 m/s

acceleration of car = 1.2 m/s²

passes the truck in 545 m

speed of the car when it just pass the truck = ?

time taken by the truck to travel 545 m

              time =\dfrac{distance}{speed}

              time =\dfrac{545}{28}

              time =19.46 s

velocity of the car when it crosses the truck

S = ut + \dfrac{1}{2}at^2

545= u\times 19.46 + \dfrac{1}{2} \times 1.2 \times 19.46^2

u = 16.33 m/s

the velocity of car when it passes the truck is u = 16.33 m/s

5 0
3 years ago
I need help PLEASE HEEEEEEEEELPP
lana66690 [7]

Answer:

a = (v2 - v1) / t

From A to B    (8 - 4) m/s / 1 s = 4 m / s^2

From A to D    ( 7 - 4) m/s / 5 s = .6 m / s^2

Note these equations hold for "uniform" values

They say nothing about the acceleration at intermediate points - the equation just says that his average speed increased from 4 m/s to 7 m/s during a 5 sec period

6 0
3 years ago
A 20-gram bullet traveling at 250 m/s strikes a block of wood that weighs 2 kg. With what velocity will the block and bullet mov
Shkiper50 [21]

Answer:

the velocity of the bullet-wood system after the collision is 2.48 m/s

Explanation:

Given;

mass of the bullet, m₀ = 20 g = 0.02 kg

velocity of the bullet, v₀ = 250 m/s

mass of the wood, m₁ = 2 kg

velocity of the wood, v₁ = 0

Let the velocity of the bullet-wood system after collision = v

Apply the principle of conservation of linear momentum to calculate the final velocity of the system;

Initial momentum = final momentum

m₀v₀ + m₁v₁ = v(m₀ + m₁)

0.02 x 250  + 2 x 0    =    v(2  + 0.02)

5 + 0 = v(2.02)

5 = 2.02v

v = 5/2.02

v = 2.48 m/s

Therefore, the velocity of the bullet-wood system after the collision is 2.48 m/s

6 0
3 years ago
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