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2 years ago

A person is in a car crash and is not wearing their seatbelt. They hit their

Physics

1 answer:

Greeley [361]2 years ago

5 0

He still move deforce stopping this call law of inertia depend on the mass of object

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A ring placed along y 2 + z 2 = 4, x = 0 carries a uniform charge of 5 μ C/m. ( a ) Find D at P(3,0,0) . ( b ) If two identical

GuDViN [60]

For right now idk :p

3 0

3 years ago

Three beads are placed on the vertices of an equilateral triangle of side d = 3.4cm. The first bead of mass m1=140gis placed on

Vlad [161]

Answer:

Xcm = 1.95 cm and Ycm = 1.76 cm

Explanation:

The very useful concept of mass center is

R cm = 1/M ∑ $m_{i}$ $r_{i}$

Where ri, mi are the mass positions of the bodies from some reference point by selecting and M is the total mass of the body.

Let's look for the total mass

M = m₁ + m₂ + m₃

M = 140 + 45 + 85

M = 270 g

Let's look for the position of each point

Point 1. top vertex, if the triangle has as side d

R₁ = d / 2 i ^ + d j ^

R₁ = (1.7 cm i ^ + 3.4 j ^) cm

Point 2. left vertex. What is the origin of the system?

R₂ = 0

Point 3. Right vertex

R₃ = d i ^

R₃ = 3.4 i ^ cm

a) The x component of the massage center

Xcm = 1 / M (m₁ x₁ + m₂ x₂ + m₃ x₃)

Xcm = 1 / M (m₁ d / 2 + 0 + m₃ d)

Xcm = d / M (m₁ / 2 + m₃)

b) Let's write the mass center component x

Xcm = 1/270 (1.7 140 + 0 + 3.4 85)

Xcm = 238/270

Xcm = 1.95 cm

c) let's find the component and center of mass

Ycm = 1 / M (m₁ y₁ + m₂ y₂ + m₃ y₃)

Ycm = 1 / M (m₁ d + 0 + 0)

Ycm = m₁ / M d

d) let's calculate

Y cm = 1/270 (140 3.4 + 0 + 0)

Ycm = 1.76 cm

8 0

3 years ago

A particle of mass m moves under an attractive central force F(r) = -Kr4 with angular momentum L. For what energy will the motio

docker41 [41]

Answer:

Angular velocity is same as frequency of oscillation in this case.

ω = $\sqrt{\frac{7K}{m} }$ x $[\frac{L^{2}}{mK}]^{3/14}$

Explanation:

- write the equation F(r) = -K $r^{4}$ with angular momentum <em>L</em>

- Get the necessary centripetal acceleration with radius r₀ and make r₀ the subject.

- Write the energy of the orbit in relative to r = 0, and solve for "E".

- Find the second derivative of effective potential to calculate the frequency of small radial oscillations. This is the effective spring constant.

- Solve for effective potential

- ω = $\sqrt{\frac{7K}{m} }$ x $[\frac{L^{2}}{mK}]^{3/14}$

3 0

3 years ago

Two boxes are at rest on a smooth, horizontal surface. The boxes are in contact with one another. If box 1 is pushed with a forc

Maslowich

Answer:

mass of box 1 = 2.20 kg

mass of box 2 = 5.93 kg

Explanation:

Let the mass of box 1 and box 2 is respectively

$m_1$ and $m_2$

so we will have

Force applied on box 1 then acceleration

$a = \frac{F}{m_1 + m_2}$

$1.50 = \frac{12.2}{m_1 + m_2}$

$m_1 + m_2 = 8.13$

Now we know that contact force between them in above case is given as

$F_n = m_2a$

$8.90 = m_2(1.5)$

$m_2 = 5.93 kg$

now we have

$m_1 = 2.20 kg$

8 0

3 years ago

a horse moves a sleigh 1.00 kilometers by applying a horizontal 2000 Newton force on its harness for 45 minutes. what is the pow

tatyana61 [14]

-- If 2,000 newtons of force were applied through a distance of 1,000 meters,
then 2,000,000 newton-meters = 2,000,000 joules of work were done.

-- 45 minutes = (45 x 60) = 2,700 seconds

-- Power = (work) / (time) = (2,000,000 j) / (2,700 s) = <u>740.74 watts</u>

Interestingly, that's almost exactly 1 horsepower. (0.99295... of 746 watts)

8 0

3 years ago