All categories

2 years ago

An object has a mass of 785 g and a volume of 15 cm³. What is its density? (Give your answer in g/cm³ to 2 decimal places).

Physics

1 answer:

Anvisha [2.4K]2 years ago

4 0

Answer:

denisity = 52.33 g/c $m^{3}$

Explanation:

Density:

$d = \frac{m}{v}$

We have that m = 785 and that v = 15 c $m^{3}$ .

$d = \frac{785}{15}$

d = 52.33 $m^{3}$

You might be interested in

Solve A and B using energy considerations.

Alisiya [41]

<span>Use the kinematic equation vf^2 = vi^2 + 2ad where;
vf = ?
vi = 0 m/s
a = 9.8 m/s^2
d1 = 10 m
d2 = 25 m

final velocity at the ground (d1): vf = sqrt(2)(9.8)(10) = 14 m/s
final velocity to the bottom of the cliff (d2): vf = sqrt(2)(9.8)(25) = 22.14 m/s
</span>

7 0

2 years ago

A 2 kg stone is tied to a 0.5 m string and swung around a circle at a constant angular velocity of 12 rad/s. the angular momentu

Illusion [34]

Starting from the angular velocity, we can calculate the tangential velocity of the stone:
$v=\omega r= (12 rad/s)(0.5 m)= 6 m/s$

Then we can calculate the angular momentum of the stone about the center of the circle, given by
$L=mvr$
where
m is the stone mass
v its tangential velocity
r is the radius of the circle, that corresponds to the length of the string.

Substituting the data of the problem, we find
$L=(2 kg)(6 m/s)(0.5 m)=6 kg m^2 s^{-1}$

3 0

2 years ago

Consider a light, single-engine airplane such as the Piper Super Cub. If the maximum gross weight of the airplane is 7780 N, the

viva [34]

Answer:

V=19.08 m/s

Explanation:

Airplane gross weight w=7780 N

Airplane wing area S=16.6 m²

Air density p=1.2250 kg/m³

Maximum lift coefficient CL=2.1

To find

Stalling Speed

Solution

The equation to find stalling speed is given below

$W=(1/2)S_{area}(V_{Stalling-speed} )^{2} (P_{Air-density} )(C_{Lmax} )\\ so\\V_{Stalling-speed}=\sqrt{\frac{2W}{S_{area}*(P_{Air-density} )(C_{Lmax} )} }\\V_{Stalling-speed}=\sqrt{\frac{2*7780}{16.6*2.1*1.225} }\\ V_{Stalling-speed}=19.08 m/s$

5 0

2 years ago

You stand on a straight desert road at night and observe a vehicle approaching. This vehicle is equipped with two small headligh

LuckyWell [14K]

For a circular aperture, the first minima (n=1) as an angular separation from the peak of the central maxima given by

Sinθ = 1.22λ / d

Where,

d is the aperture or pupil diameter

d = 4.69 mm = 4.69 × 10^-3m

λ is the wavelength

λ = 545 nm = 545 × 10^-9 m

Then,

Sinθ = 1.22λ / d

Sinθ = 1.22 × 545 × 10^-9 / 4.69 × 10^-3

Sinθ = 1.418 × 10^-4 rad

Then, the head light sources have the same angular separation θ from the eye as the image have inside the eye.

For the headlight

Sinθ ≈ light separation / distantce for the eye

Light separation is give as x = 0.659 m

And let the distance of the eye be D

Then,

Sinθ = x / D

Make D subject of formula

D = x / Sinθ

D = 0.695 / 1.418 × 10^-4

D = 4902.316m

To km, 1km = 1000m

D ≈ 4.9 km

4 0

2 years ago

Jeff puts on a leather jacket over his sweater. The sweater becomes negatively charged. Which statements about Jeff’s situation

nikdorinn [45]

b. The sweater has a tendency to attract protons.

8 0

2 years ago

Read 2 more answers