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4 years ago

An object has a mass of 14 grams and a density of 7 g/mL. When placed in water it sinks. What is the Volume of water displaced?

Physics

1 answer:

iVinArrow [24]4 years ago

3 0

Explanation:

D = M/V

We rearrange this equation to get V = M/D

So 14 g/7 g/mL,

the grams cancel out, and we're left with 2 mL.

D is the answer.

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Anna drew a diagram to compare the strong and weak force. Which labels belong in the areas marked X, Y, and Z? X: infinite range

Maslowich

Answer:

For areas marked X, Y, Z, X is attractive only, Y has a very small range, and Z is attractive and repulsive

Explanation:

Solution

Given that:

From the question stated, Anna drew a diagram to compare forces that are strong and weak.

Now,

We are to find which labels are grouped in areas marked as X, Y, Z respectively.

Thus,

For X, Y, Z it is marked as:

X: Always attractive or attractive only

Y: Very small range

Z: Repulsive and attractive

8 0

3 years ago

The radius of curvature of both sides of a converging lens is 18 cm. One side of the lens is coated withsilver so that the inner

Dahasolnce [82]

Answer:

n = 1.4

Explanation:

Given,

R1 = 18 cm, R2 = -18 cm

From lens makers formula

1/f = (n - 1)(1/18 + 1/18) = (n-1)/9

f = 9/(n-1)

Power, P = 1/f ( in m) = (n-1)/0.09

Now, this lens is in with conjunction with a concave mirror which then can be thought of as to be in conjunction with another thin lens

Power of concave mirror = P' = 1/f ( in m) = 2/R = 2/0.18 = 1/0.09

Net power of the combination = 2P + P' = 2(n-1)/0.09 + 1/0.09 = 1/0.05

n = 1.4

7 0

3 years ago

What is your acceleration near earth due to gravity

Maksim231197 [3]

I dont know but i know i dont lnow if this is true but the gravity is slowly going away every 4 year i dont know i think.

7 0

3 years ago

A dentist’s drill starts from rest. After 4.28 s of constant angular acceleration it turns at a rate of 28940 rev/min. Find the

mylen [45]

Answer:

Angular acceleration, is $708.07\ rad/s^2$

Explanation:

Given that,

Initial speed of the drill, $\omega_i=0$

After 4.28 s of constant angular acceleration it turns at a rate of 28940 rev/min, final angular speed, $\omega_f=28940\ rev/min=3030.58\ rad/s$

We need to find the drill’s angular acceleration. It is given by the rate of change of angular velocity.

$\alpha =\dfrac{\omega_f}{t}\\\\\alpha =\dfrac{3030.58\ rad/s}{4.28\ s}\\\\\alpha =708.07\ rad/s^2$

So, the drill's angular acceleration is $708.07\ rad/s^2$ .

4 0

3 years ago

Influenced by the gravitational pull of a distant star, the velocity of an asteroid changes from from +19.3 km/s to −18.8 km/s o

muminat

As per above given data

initial velocity = 19.3 km/s

final velocity = - 18.8 km/s

now in order to find the change in velocity

$\Delta v = v_f - v_i$

$\Delta v = -18.8 - 19.3$

$\Delta v = -38.1 km/s$

$\Delta v = -3.81 * 10^4 m/s$

Part b)

Now we need to find acceleration

acceleration is given by formula

$a = \frac{\Delta v}{\Delta t}$

given that

$\Delta v =- 3.81 * 10^4 m/s$

$\Delta t = 2.07 years = 6.53 * 10^7 s$

now the acceleration is given as

$a = \frac{-3.81 * 10^4}{6.53 * 10^7}$

$a = - 5.84 * 10^{-4}m/s^2$

so above is the acceleration

4 0

3 years ago