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2 years ago

A uniform rod of length L, mass M, is suspended by two thin strings. Which of the following statements is true

Physics

1 answer:

Fantom [35]2 years ago

6 0

Answer:

(d) not enough info

Explanation:

because it doesn't specify where the strings are attached

if it was the two ends of the rod then T1 would be equal to T2

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A cube of wood 15.0 cm on each side is tied to the bottom of a tank filled with water to a depth of 50 cm. The tension in the st

Elina [12.6K]

Answer:

(a). The density of the wood is $1479.48\times10^{2}\ Kg/m^3$

(b). The absolute pressure at the bottom of the tank is $1.06200\times10^{5}\ Pa$ .

Explanation:

Given that,

Side of cube = 15.0 cm

Depth = 50 cm

Tension = 6.615 N

We need to calculate the volume of the wood

Using formula of volume

$V = a^3$

$V=(15.0\times10^{-2})^3$

$V=0.003375\ m^3$

We need to calculate the density of the wood

Using buoyant force

$\rho_{w}gh=mg+T$

$\rho_{w}gh=\rho_{c}Vg+T$

Put the value into the formula

$\rho_{c}=\dfrac{\rho_{w}gh-T}{Vg}$

Put the value into the formula

$\rho_{c}=\dfrac{1000\times9.8\times50\times10^{-2}-6.615}{0.003375\times9.8}$

$\rho_{c}=1479.48\times10^{2}\ Kg/m^3$

(b). We need to calculate the absolute pressure at the bottom of the tank

Using formula of absolute pressure

$P=P_{atm}+\rho gh$

Put the value into the formula

$P=1.013\times10^{5}+1000\times9.8\times0.5$

$P=1.06200\times10^{5}\ Pa$

Hence, (a). The density of the wood is $1479.48\times10^{2}\ Kg/m^3$

(b). The absolute pressure at the bottom of the tank is $1.06200\times10^{5}\ Pa$ .

7 0

2 years ago

ANSWER ASAP which of the following is true about cells?

GalinKa [24]

C is the correct answer

3 0

2 years ago

Read 2 more answers

A charge of +3.00 μC is located at the origin, and a second charge of -2.00 μC is located on the x-y plane at the point (50.0 cm

romanna [79]

Answer:

0.0567 N

Explanation:

q1 = 3 micro coulomb

q2 = - 2 micro coulomb

OB = 50 cm

AB = 60 cm

By using Pythagoras theorem in triangle OAB

$OA^{2}=OB^{2}+AB^{2}$

$OA^{2}=50^{2}+60^{2}=6100$

OA = 78.1 cm

By using the Coulomb's law, the force on 3 micro coulomb due to -2 micro coulomb is

$F=\frac {Kq_{1}q_{2}}{OA^{2}}= \frac{9 \times 10^{9}}\times 3\times 10^{-6} \times 2 \times 10^{-6}{0.781^{2}}$

F = 0.0885 N

The horizontal component of force is

= F CosФ = $F\times \frac{OB}{OA}$

= 0.0885 x 50 / 78.1 = 0.0567 N

8 0

2 years ago

A golf ball is hit so that it leaves the club face at a velocity of 45m/s at an angle of 40° to the horizontal. by ignoring the

tatuchka [14]

Answer:a)45cos40 b)45sin40 c) 2.95 d)102

Explanation:

a) $v_{x} = 45 cos40$

b) $v_{y}=45sin40$

c) Since air resistance is negligible, the only force acting on the golf ball is gravity. Thus, its vertical acceleration is -<em>g</em>. We know the final velocity must be 0 m/s, because this will be when the golf ball reaches the maximum height and starts to change direction (it falls back to the ground). We also know initial velocity in the vertical direction (see part b). Thus, we can use this equation: $v_{f} = v_{i} + at$ .

$0 m/s = 45sin40 + (-9.8m/s^2)t\\t = 2.95s$

d) The horizontal distance traveled is dependent on (1) how long the ball is in the air and (2) what the horizontal velocity is. (1) was found in part c, and (2) was found in part a.

$x=vt\\x=45cos40(2.95) =102m$

6 0

1 year ago

Explain why a book placed on a table does not move?

kifflom [539]

Answer:

The force that the book exerts down on the table is equal to the force the table exerts upward.

Explanation:

I am 100% positive. That results in balanced forces which is why it does not move. If you did want it to move, you would apply unbalanced forces.

4 0

2 years ago

Read 2 more answers