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3 years ago

true or false:acceleration toward the center of a curved or circular path is called gravitational acceleration.

1 answer:

5 0

Nope. It's called 'centripetal' acceleration. The force that created it MAY be gravitational, but it doesn't have to be. For things on the surface of the Earth moving in circles, it's never gravity.

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A rock, with a density of 3.55 g/cm^3 and a volume of 470 cm^3, is thrown in a lake. a) What is the weight of the rock out of th

SIZIF [17.4K]

Answer:

a) Weight of the rock out of the water = 16.37 N

b) Buoyancy force = 4.61 N

c) Mass of the water displaced = 0.47 kg

d) Weight of rock under water = 11.76 N

Explanation:

a) Mass of the rock out of the water = Volume x Density

Volume = 470 cm³

Density = 3.55 g/cm³

Mass = 470 x 3.55 = 1668.5 g = 1.6685 kg

Weight of the rock out of the water = 1.6685 x 9.81 = 16.37 N

b) Buoyancy force = Volume x Density of liquid x Acceleration due to gravity.

Volume = 470 cm³

Density of liquid = 1 g/cm³

$\texttt{Buoyancy force}= \frac{470\times 1\times 9.81}{1000} = 4.61 N$

c) Mass of the water displaced = Volume of body x Density of liquid

Mass of the water displaced = 470 x 1 = 470 g = 0.47 kg

d) Weight of rock under water = Weight of the rock out of the water - Buoyancy force

Weight of rock under water = 16.37 - 4.61 =11.76 N

3 0

3 years ago

Even when shut down after a period of normal use, a large commercial nuclear reactor transfers thermal energy at the rate of 150

Sever21 [200]

Answer: $2.89\approx 2.9^{\circ}C/s$

Explanation:

Given

$Power\left ( P\right )=150 MW$

mass of core $\left ( m\right )=1.60\times 10^5 kg$

Average specific heat $\left ( C\right )=0.3349 KJ/kg^{\circ}C$

And rate of increase of temperature = $\frac{\mathrm{d}T}{\mathrm{d} t}$

Now

P= $mc\frac{\mathrm{d}T}{\mathrm{d} t}$

$150\times 10^6=1.60\times 10^5\times 0.3349\times \frac{\mathrm{d}T}{\mathrm{d} t}$

Thus $\frac{\mathrm{d}T}{\mathrm{d} t}=[tex]\frac{1.60\times 10^5\times 0.3349}{150\times 10^6}$

$\frac{\mathrm{d}T}{\mathrm{d} t}=2.89\approx 2.9^{\circ}C/s$

6 0

3 years ago

An organ pipe open at both ends has two successive harmonics with frequencies of 220 Hz and 240 Hz. What is the length of the pi

Harman [31]

Answer:

The length is $l = 8.6 \ m$

Explanation:

From the question we are told that

The frequencies of the two successive harmonics are $f_1 = 220 \ Hz$ , $f_2 = 240 \ Hz$

The speed of sound in the air is $v_s = 343 \ m/s$

Generally the frequency of a given harmonic is mathematically represented as

$f_n = \frac{n v }{2l}$

Here n defines the position of the harmonics

Now since the position of both harmonic is not know but we know that they successive then we can represented them mathematically as

$220 = \frac{n v}{2l}$

and

$240 = \frac{(n+1) v}{2l}$

$\frac{(n + 1 ) v}{2l} - \frac{n v}{2l} = 240-220$

=> $\frac{v}{2l} = 20$

=> $l = 8.6 \ m$

7 0

3 years ago

Suppose an object is moving in a straight line at 50 miles/hr. According to Newton's first law of motion, the object will ______

Setler [38]

Answer:

Explanation:

Newton's first law of motion states that a body will remain in its state of rest or if its in motion will continue to move in a straight line, unless its acted upon by an external force.The ability of an object to stay at rest or in motion if its in motion is known as inertia.

Hence the correct option is B.

8 0

3 years ago

Determine the amount of work done by the applied force when a 87 N force is applied to move a 15 kg object a horizontal

elena-s [515]

Answer:

391.5 J

Explanation:

The amount of work done can be calculated using the formula:

W = F║d
where the force is parallel to the displacement

Looking at the formula, we can see that the mass of the object does not affect the work done on it.

Substitute the force applied and the displacement of the object into the equation.

W = (87 N)(4.5 m)
W = 391.5 J

The amount of work done on the object is 391.5 J in order to move it 4.5 meters with an applied force of 87 Newtons.

5 0

2 years ago

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