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Rudiy27
2 years ago
15

If a wood is drowned in water up to 2/3 of its volume and the same wood is dipped in a oil upto 9/10 of its volume.What's the de

nsity of wood and the relative oil?
Density of water -1000kgm-3
Physics
1 answer:
MrMuchimi2 years ago
7 0
S.G. or specific gravity is the ratio of the density of an object to a fluid. It also is the fraction of the object's volume that would be submerged if the object were placed in the fluid. (note: if S.G.>1, the object will sink)

S.G. = density(wood) / density(water)

2/3 = density(wood) / 1000

density(wood) = 2000/3 kg/m^3 = ~666.67 kg/m^3

S.G. = density(wood) / density(oil)

9/10 = 2000/3 / density(oil)

density(oil) = 2000/3 / 9/10 = 20000/27 kg/m^3 = ~740.74 kg/m^3
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To form a negative ion, an atom will give up an electron to gain stability <br><br> True or False?
Mumz [18]
It is a completely false statement that to form a negative ion, an atom will <span>give up an electron to gain stability. The correct option among all the options that are given in the question is the second option. Negative ion has more electrons and so it would be wrong to say that it will give up an electron. I hope the answer has helped you.
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3 0
2 years ago
3) A driver in a 1000-kg car traveling at 24 m/s slams on the brakes and skids to a stop. If the coefticient of friction between
Natali5045456 [20]

Answer:

A) 37 m

Explanation:

The car is moving of uniformly accelerated motion, so the distance it covers can be calculated by using the following SUVAT equation:

v^2 -u^2 = 2ad (1)

where

v = 0 m/s is the final velocity of the car

u = 24 m/s is the initial velocity

a is the acceleration

d is the length of the skid

We need to find the acceleration first. We know that the force responsible for the (de)celeration is the force of friction, so:

F=ma=-\mu mg

where

m = 1000 kg is the mass of the car

\mu = 0.80 is the coefficient of friction

a is the deceleration of the car

g = 9.8 m/s^2 is the acceleration due to gravity

The negative sign is due to the fact that the force of friction is against the motion of the car, so the sign of the acceleration will be negative because the car is slowing down. From this equation, we find:

a=-\mu g

And we can substitute it into eq.(1) to find d:

v^2 -u^2 = 2(\-mu g) d\\d= \frac{v^2-u^2}{-2 \mu g}=\frac{0-(24 m/s)^2}{-2(0.80)(9.8 m/s^2)}=36.7 m \sim 37 m

7 0
3 years ago
What is the spring constant for a supermarket scale that stretches 0.01 m when a force of 4 N is applied
bija089 [108]

Answer:

400N/m

Explanation:

K = F/x

= 4/0.01

= 400 N/m

5 0
2 years ago
A baseball bat hits a baseball with a force of 100 newtons. What is the force and its direction exerted by the ball on the bat?
Brut [27]

Answer:

The force exerted by the ball on the bat has a magnitude of 100 N and its direction is exactly opposite to that of the force exerted by the bat on the ball.

Explanation:

Recall that Newton's third law tells us that : "For every action, there is an equal and opposite reaction."

Therefore if the bat acts on the ball with a force of 100 N, the ball acts on the bat with a similar magnitude of force (100 N) but direction opposite to the original force.

F_{ab} = -\,F_{ba}

7 0
3 years ago
A toaster using a Nichrome heating element operates on 120 V. When it is switched on at 28 ∘С, the heating element carries an in
sukhopar [10]

Answer:

The final temperature of the element = 262.67°C

The power dissipated in the heating element initially = 163.21 W

The power dissipated in the heating element when the current reaches 1.23 A = 147.60 W

Explanation:

Our given parameters include;

A Nichrome heating element operates on 120 V.

Voltage (V) = 120V

Initial Current (I₁) = 1.36 A

Initial Temperature (T₁) = 28°C

Final Current (I₂) = 1.23 A

Final Temperature (T₂) = unknown ????

Temperature dependencies of resistance is given by:

R_{T(2)}=R_1[1+\alpha (T_2-T_1)]            ----------------------    (1)

in which R₁ is the resistance at temperature T₁

R_{T(2) is the resistance at temperature T₂

Given that V= IR

R = \frac{V}{I}

Therefore, the resistance at temperature 28°C is;

R_{28}= \frac{120V}{1.36A}

= 88.24Ω

R_{T(2) = \frac{120V}{1.23A}

= 97.56Ω

From (1) above;

R_{T(2)}=R_1[1+\alpha (T_2-T_1)]      

97.56 = 88.24 [ 1 + 4.5×10⁻⁴(°C)⁻¹(T₂-28°C)]

\frac{97.56}{88.24}= 1+(4.5*10^{-4})(T-28^0C)

1.1056 - 1 = 4.5×10⁻⁴(°C)⁻¹(T₂-28°C)

0.1056 = 4.5×10⁻⁴(T₂-28°C)

\frac{0.1056}{4.5*10^{-4}}= T-28^0C

T - 28° C = 234.67

T = 234.67 + 28° C

T = 262.67 ° C

(b)

What is the power dissipated in the heating element initially and when the current reaches 1.23 A

The power dissipated in the heating element initially can be calculated as:

P = I²₁R₂₈

P = (1.36A)²(88.24Ω)

P = 163.209 W

P ≅ 163.21 W

The power dissipated in the heating element when the current reaches 1.23 A can be calculated as:

P= I^2_2R_{T^0C

P = (1.23)²(97.56Ω)

P = 147.598524

P ≅ 147.60 W

6 0
2 years ago
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