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pshichka
1 month ago
9

a 62.0 kg curler runs into a stationary 78.1 kg curler and they hold on to each other. Together they move away at 1.29 m/s, west

. What was the original velocity of the 62.0 kg curler?
Physics
1 answer:
Maru [3.3K]1 month ago
7 0
Momentum P can be expressed as:
P = mv

The inputs are: m₁ = 62 kg, m₂ = 78.1 kg, v₁ =?, v₂ = 0 m/s, and v₁₊₂ = 1.29 m/s

Before:
P = m_1 v_1 * m_2 v_2 = m_1 v_1

After:
P = (m_1 + m_2)v_{1+2}

As momentum is conserved, the momentum prior to the collision must equal the momentum after the collision.

P = m_1 v_1 = (m_1 + m_2)v_{1+2} \\ v_1 = \frac{(m_1 + m_2)v_{1+2}}{m_1}

To solve for v₁.
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A bowling pin is thrown vertically upward such that it rotates as it moves through the air, as shown in the figure. Initially, t
Ostrovityanka [3204]

Answer:

Explanation:

The equation used to determine the maximum height of the bowling pin during its trajectory is given by;

H = u²/2g

where u, the initial speed/velocity, equals 10m/s

g stands for gravitational acceleration = 9.81m/s²

Substituting in the values gives us

H = 10²/2(9.81)

H = 100/19.62

Consequently, the highest point of the bowling pin's center of mass is approximately 5.0m.

3 0
2 months ago
Use the formula t = (0.25) s1/2 to find the time t in seconds it will take a stone to drop a distance s of 200 feet. Round your
inna [3103]

Answer:

The duration, t = 3.53 seconds

Explanation:

The following information is provided:

The equation to calculate the time t is expressed as:

t=(0.25)s^{1/2}...... (1)

Where

s denotes the distance in feet

We are to determine the duration taken by the stone to fall a distance of 200 feet, where s = 200 feet

Substituting the value of s into equation (1) yields:

t=(0.25)\times (200)^{1/2}

t = 3.53 seconds

Thus, the time taken by the object is 3.53 seconds, which provides the required answer.

4 0
2 months ago
A 1.97-pF capacitor with a plate area of 5.86 cm2 and separation between the plates of 2.63 mm is connected to a 9.0-V battery a
inna [3103]

If the plate separation is modified after the battery is disconnected, the updated distance between plates is 9.21 mm

If changes are made while the battery remains connected, the new separation becomes 0.11 mm

The capacitance for an air-filled parallel plate capacitor can be expressed as:

C=\frac{\epsilon_0A}{d}

In this equation, \epsilon_0 refers to the permittivity of free space, A stands for the plate area, and D represents the separation distance.

Thus,

C \alpha \frac{1}{d}.......(1)

Therefore, should the distance between the plates shift from d₁ to d₂, the capacitance ratio in both scenarios can be represented as:

\frac{C_1}{C_2} =\frac{d_2}{d_1}......(2)

Scenario (i)

When the capacitor is fully charged and then disconnected from the battery before adjusting the plate distance, the charge will remain steady while the capacitance varies.

The initial energy E₁ stored in the capacitor can be expressed as:

E_1=\frac{Q^2}{2C_1}......(3)

Once the separation changes to d₂, capacitance becomes C₂, but the charge Q remains unchanged.

Thus,

E_2=\frac{Q^2}{2C_2}......(4)

By dividing equation (4) by (3),

\frac{E_2}{E_1} =\frac{C_1}{C_2}

According to equation (2),

\frac{E_2}{E_1} =\frac{C_1}{C_2}=\frac{d_2}{d_1}

This results in a 3.5 fold increase in energy.

\frac{E_2}{E_1} =\frac{d_2}{d_1}=3.5\\ d_2=3.5*2.63 mm\\ =9.205 mm=9.21 mm

Scenario (2)

If the capacitor is kept connected to the power source, the voltage V across the plates will remain unchanged.

The initial energy is described as

E_1=\frac{1}{2} C_1V^2......(5)

The final energy when the plate separation transitions to d₂ can be written as:

E_2=\frac{1}{2} C_2V^2.....(6)

Referencing equations (5) and (6)

\frac{E_2}{E_1} =\frac{C_2}{C_1}

From equation (2),

\frac{E_2}{E_1} =\frac{C_2}{C_1}=\frac{d_1}{d_2}

Thus, in this particular scenario,

\frac{E_2}{E_1} =\frac{d_1}{d_2}\\d_2=\frac{d_1}{3.5} \\ =\frac{2.63 mm}{3.5} \\ =0.109 mm=0.11 mm

Therefore,

Adjusting plate separation after battery disconnection yields 9.21 mm

If modified while connected, the new separation measures 0.11 mm





6 0
1 month ago
A plane flying at 70.0 m/s suddenly stalls. If the acceleration during the stall is 9.8 m/s2 directly downward, the stall lasts
ValentinkaMS [3465]

Answer:

v = 66.4 m/s

Explanation:

We know that the aircraft starts off moving at a speed of

v = 70 m/s

now we have

v_x = 70 cos25

v_x = 63.44 m/s

v_y = 70 sin25

v_y = 29.6 m/s

in the Y direction, we can apply kinematic equations

v_y = v_i + at

v_y = 29.6 - (9.81 \times 5)

v_y = -19.5 m/s

as there is no acceleration along the x-axis, the velocity in this direction remains unchanged

thus yielding

v = \sqrt{v_x^2 + v_y^2}

v = \sqrt{63.44^2 + 19.5^2}

v = 66.4 m/s

4 0
2 months ago
The nine ring wraiths want to fly from barad-dur to rivendell. rivendell is directly north of barad-dur. the dark tower reports
kicyunya [3294]
To counteract a 58 mph crosswind, the western component of the trajectory must be accounted for. Consequently, directing towards the northwest creates a 45-degree angle, aligning with the destination. This triangle's third vertex is located at the destination, with the right angle positioned there. The western aspect of their flight represents the triangle's base, while the vertical side reflects the resultant path, and the hypotenuse indicates the actual distance traveled. Since the 58 mph crosswind was countered by flying in a northwest direction, the distance from the starting point to the destination should equal the westward segment of their journey. The hypotenuse can be determined via the square root of twice the dimension of the identical sides.
c = sqrt (58^2 + 58^2) = sqrt (6728) = 82.02

An alternative method:

c = sqrt (2) * 58 = 1.414 * 58 = 82.02

Thus, they must fly at 82.02 mph.
5 0
1 month ago
Read 2 more answers
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