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3 years ago

6

A scientist pours 0.120 L of solution A into an Erlenmeyer flask. She adds 2.345 L of solution B. How many significant figures a

re in the resulting answer?

1 answer:

nika2105 [10]3 years ago

6 0

0.120L + 2.345L = 2.465L = 4 significant figures in the answer

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20 points<br> Which BEST replaces the question mark?<br><br> In Saudi Arabia, __________?

dsp73

<span>The economy is based upon the production of oil. Saudi Arabi is one of the leading oil producing countries in the world. It is one of the founding counties of Opec, an oil cartel made up of middle eastern oil producing countries which try to manipulate the worldwide price of oil.</span>

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4 years ago

two test cars of equal mass moving towards each other collide on a horiontal frictionless surface. Before the collision, car A h

LiRa [457]

Answer:

4 m/s

Explanation:

m1 = m2 = m

u1 = 20 m/s, u2 = - 12 m/s

Let the speed of composite body is v after the collision.

Use the conservation of momentum

Momentum before collision = momentum after collision

m1 x u1 + m2 x u2 = (m1 + m2) x v

m x 20 - m x 12 = (m + m) x v

20 - 12 = 2 v

8 = 2 v

v = 4 m/s

Thus, the speed of teh composite body is 4 m/s.

4 0

3 years ago

Alex is asked to move two boxes of books in contact with each other and resting on a rough floor. He decides to move them at the

Over [174]

Answer:

Part a)

$a= 0.32 m/s^2$

Part b)

$F_c = 3.6 N$

Part c)

$F_c = 5.5 N$

Explanation:

Part a)

As we know that the friction force on two boxes is given as

$F_f = \mu m_a g + \mu m_b g$

$F_f = 0.02(10.6 + 7)9.81$

$F_f = 3.45 N$

Now we know by Newton's II law

$F_{net} = ma$

so we have

$F_p - F_f = (m_a + m_b) a$

$9.1 - 3.45 = (10.6 + 7) a$

$a = \frac{5.65}{17.6}$

$a= 0.32 m/s^2$

Part b)

For block B we know that net force on it will push it forward with same acceleration so we have

$F_c - F_f = m_b a$

$F_c = \mu m_b g + m_b a$

$F_c = 0.02(7)(9.8) + 7(0.32)$

$F_c = 3.6 N$

Part c)

If Alex push from other side then also the acceleration will be same

So for box B we can say that Net force is given as

$F_p - F_f - F_c = m_b a$

$9.1 - 0.02(7)(9.8) - F_c = 7(0.32)$

$F_c = 9.1 - 0.02 (7)(9.8) - 7(0.32)$

$F_c = 5.5 N$

3 0

3 years ago

A nature photographer is using a camera that has a lens with a focal length of 3.06 cm. The photographer is taking pictures of a

sleet_krkn [62]

Answer:

film is at distance of 3.07 cm from lens

Explanation:

Given data

focal length = 3.06 cm

distance = 10.4 m = 1040 cm

to find out

How far must the lens

solution

we apply here lens formula that is

1/f = 1/p + 1/q

here f = 3.06 and p = 1040 so we find q

1/f = 1/p + 1/q

1/3.06 = 1/1040 + 1/q

1/ q = 0.3258

q = 3.0690 cm

so film is at distance of 3.07 cm from lens

6 0

4 years ago

Could you help me with this question?

Scorpion4ik [409]

Answer:

Magnitude of static friction force is 70 sin40° = 44.99 N.

No, it is not necessary that it is maximum static friction.

Normal force is equal to 70 cos40° = 53.62 N.

Explanation:

We apply newton law of moton equation along the plane and perpendicular to plane;

Along the plane,

70 sin 40° = $f_{r}$ ---------------(1)

70 cos 40° = N --------------(2)

$f_{r}_{max}$ = μN -----------------(3)

So, it depends on the value of μ that the friction is maximum or not .

7 0

3 years ago