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3 years ago

Read the sentence.

Physics

2 answers:

irakobra [83]3 years ago

6 0

Answer:

The answer is C (Assembled)

Explanation:

Aleksandr [31]3 years ago

5 0

Answer:

assembled

Explanation: because it was last weekend it was in the past so the past tense of assembled must be used

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A 4.00 kg block is pushed along the ceiling with a constant applied force of 85.0 N that acts at an angle of 55.0 degrees with t

steposvetlana [31]

Answer:

0.35

Explanation:

According to Newton's second law;

\sum Fx = ma

Fm - Ff =ma

Fm is the moving force = Wsin theta

Fm = 4(9.8)sin55

Fm = 32.1N

Ff is the frictional force = nmgcos theta

Ff = n(4)(9.8)cos55

Ff = 22.48n

Acceleration a = 6.0m/s²

Substitute the given values into the formula and get the coefficient of friction

32.11-23.48n = 4(6)

32.11-24= 23.48n

8.11 = 23.48

n = 8.11/23.48

n = 0.35

Hence the coefficient of friction is 0.35

6 0

3 years ago

Which of the following, by scientific definition is NOT work? lifting boxes of books, pedaling on your bike, holding a book over

Karolina [17]

Answer:

holding a book over your head

3 0

3 years ago

Why are multiple images seen when two plane mirrors are placed at an angle

Whitepunk [10]

Answer: Can I get a picture???

8 0

3 years ago

Mation about your sour

Ann [662]

(#1). (D).

(#2). (C).

8 0

3 years ago

A professional racecar driver buys a car that can accelerate at 5.9 m/s2. The racer decides to race against another driver in a

3241004551 [841]

Answer:

(a) Time will be t = 3.56 sec

(b) Distance traveled by car when they are side by side is 37.38712 m

(b) Velocity of race car = 21.004 m/sec

velocity of stock car = 12.816 m/sec

Explanation:

We have given acceleration of the car $a_1=5.9m/sec^2$

Acceleration of the stock car $a_2=3.6m/sec^2$

When 1st car overtakes the second car then distance traveled by both the car will be same

(a) So $s_1=s_2$

As both car starts from rest so initial velocity of both car will be 0 m/sec

It is given that stock car leaves 1 sec before

So $\frac{1}{2}\times 5.9\times t^2=\frac{1}{2}\times (t+1)^2\times 3.6$

After solving t = 3.56 sec

(b) From second equation of motion $s=ut+\frac{1}{2}at^2=0\times 3.56+\frac{1}{2}\times 5.9\times 3.56^2=37.38712m$

So velocity of race car v = 0+5.9×3.56 = 21.004 m/sec

Velocity of stock car v = 0+ 3.6×3.56 = 12.816 m/sec

3 0

3 years ago