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2 years ago

The image of an object in a flat mirror is always the same _____ as the object.

Physics

1 answer:

ahrayia [7]2 years ago

4 0

Size. The answer is size.

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4. The time it takes an object to rotate once is

pshichka [43]

Answer:

Explanation:

23.9344696 hours

3 0

2 years ago

Which has a higher frequency, a wave that makes 7 cycles per second or one that makes 12 cycles per second?

brilliants [131]

Answer:

A sound wave that has a higher frequency is a wave that makes 12 cycles per second.

Explanation:

The frequency of a wave is the same as the frequency of the vibrations that caused the wave. This takes more energy, so a higher-frequency wave has more energy than a lower-frequency wave with the same amplitude.

6 0

2 years ago

. Find the buoyant force exerted on a ball with radius of 2 meters, when the ball is entirely immersed in the water.

almond37 [142]

Explanation:

Buoyancy force is equal to the weight of the displaced fluid:

B = ρVg

where ρ is the density of the fluid,

V is the volume of the displaced fluid,

and g is the acceleration due to gravity.

The fluid is water, so ρ = 1000 kg/m³.

The volume displaced is that of a sphere with radius 2 m:

V = 4/3 π r³

V = 4/3 π (2 m)³

V ≈ 33.5 m³

The buoyancy force is therefore:

B = (1000 kg/m³) (33.5 m³) (9.8 m/s²)

B ≈ 328,400 N

Round as needed.

8 0

2 years ago

Three point charges have equal magnitudes, two being positive and one negative. These charges are fixed to the corners of an equ

lesantik [10]

Answer:

The magnitude of the force on positive charges will be $\bf{227.06~N}$ and the magnitude of the force on the negative charge is $\bf{302.7~N}$ .

Explanation:

Given:

The value of the charges, $q = 3.5~\mu C$ .

The length of each side of the triangle, $l = 2.9~cm$ .

Consider a equilateral triangle $\bigtriangleup ABC$ , as shown in the figure. Let two point charges of magnitude $q$ are situated at points $A$ and $B$ and another point charge $-q$ is situated at point $C$ .

The value of the force on the charge at point $A$ due to charge at point $C$ is given by

$F_{CA} = \dfrac{kq^{2}}{l^{2}},~~along~CA$

The value of the force on the charge at point $A$ due to charge at point $B$ is given by

$F_{BA} = \dfrac{kq^{2}}{l^{2}},~~along~BA$

The net resultant force on the charge at point $A$ is given by

$~~~~F_{A} = \sqrt{F_{BA}^{2} + F_{CA}^{2} + 2F_{BA}F_{CA}\cos 60^{0}}\\~~~~~= \dfrac{kq^{2}}{l^{2}}\sqrt{2(1 + \cos 60^{0})}\\or, F_{A}= \dfrac{\sqrt{3}kq^{2}}{l^{2}}~~~~~~~~~~~~~~~~~~~~(1)$

The value of the force on the charge at point $B$ due to charge at point $C$ is given by

$F_{CB} = \dfrac{kq^{2}}{l^{2}},~~along~CB$

The value of the force on the charge at point $B$ due to charge at point $A$ is given by

$F_{AB} = \dfrac{kq^{2}}{l^{2}},~~along~AB$

The net resultant force on the charge at point $B$ is given by

$~~~~F_{B} = \sqrt{F_{AB}^{2} + F_{CB}^{2} + 2F_{AB}F_{CB}\cos 60^{0}}\\~~~~~= \dfrac{kq^{2}}{l^{2}}\sqrt{2(1 + \cos 60^{0})}\\or, F_{B}= \dfrac{\sqrt{3}kq^{2}}{l^{2}}~~~~~~~~~~~~~~~~~~~~(2)$

The value of the force on the charge at point $C$ due to charge at point $A$ is given by

$F_{AC} = \dfrac{kq^{2}}{l^{2}},~~along~AC$

The value of the force on the charge at point $C$ due to charge at point $B$ is given by

$F_{BC} = \dfrac{kq^{2}}{l^{2}},~~along~BC$

The net resultant force on the charge at point $C$ is given by

$F_{C} = 2F_{BC} \sin 60^{0}~~along~the~line~perpendicular~to~AB\\~~~~~= \dfrac{2kq^{2}}{l^{2}}\sin 60^{0}~~~~~~~~~~~~~~~~~~~~(3)$

Substitute $3.5~\mu C$ for $q$ , $0.029~m$ for $l$ and $9 \times 10^{9}~Nm^{2}C^{-2}$ for $k$ in equation (1), we have

$F_{A} = \dfrac{\sqrt{3}(9 \times 10^{9}~Nm^{2}C^{-2})(3.5 \times 10^{-6}~C)^{2}}{(0.029~m)^{2}}\\~~~~~= 227.06~N$

Substitute $3.5~\mu C$ for $q$ , $0.029~m$ for $l$ and $9 \times 10^{9}~Nm^{2}C^{-2}$ for $k$ in equation (2), we have

$F_{B} = \dfrac{\sqrt{3}(9 \times 10^{9}~Nm^{2}C^{-2})(3.5 \times 10^{-6}~C)^{2}}{(0.029~m)^{2}}\\~~~~~= 227.06~N$

Substitute $3.5~\mu C$ for $q$ , $0.029~m$ for $l$ and $9 \times 10^{9}~Nm^{2}C^{-2}$ for $k$ in equation (3), we have

$F_{C} = \dfrac{2(9 \times 10^{9}~Nm^{2}C^{-2})(3.5 \times 10^{-6}~C)^{2}}{(0.029~m)^{2}} \sin 60^{0}\\~~~~~= 302.7~N$

8 0

2 years ago

A 2 kg ball travellng to the right with a speed of 4 m/s collidees with a 5 kg ball traveling to the left with a speed of 3 m/s.

Savatey [412]

Explanation:

Momentum before collision:

(2 kg) (4 m/s) + (5 kg) (-3 m/s) = -3 kg m/s

No external forces act on the balls, so momentum is conserved. Therefore, momentum after collision is also -3 kg m/s.

5 0

2 years ago