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svlad2 [7]
2 years ago
6

While the cart is moving along an aisle, it comes in contact with a smear of margarine that had recently been dropped on the flo

or. Suddenly the friction force is reduced for -40.0 newtons to -20.0 newtons. What is the net force on the cart if the "pushing force" remains at 40.0 newtons? Does the grocery cart move at constant velocity over the spilled maragine?
Physics
1 answer:
liberstina [14]2 years ago
3 0

Answer:

No, the velocity of the grocery cart will not remain constant, but instead will increase continuously

Explanation:

No the grocery cart will not move at constant velocity over the spilled maragine because as the frictional force is changed to - 20 N, the net force will not be equal to zero

Net force on the cart when frictional force is changed to - 20 N is 40 - 20 N = 20 N

As the force is positive, it means that the force is acting in the direction of motion of the cart and will increase the velocity of the cart as the force is acting in the direction of motion

∴ The velocity of the cart will not remain constant instead it will increase

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A particular cylindrical bucket has a height of 36.0 cm, and the radius of its circular cross-section is 15 cm. The bucket is em
Sergeeva-Olga [200]

Answer:

a. 0.000002 m

b. 0.00000182 m

Explanation:

36 cm = 0.36 m

15 cm = 0.15 m

a) We can start by calculating the air-water pressure of the bucket submerged 20m below the water surface:

P = \ro g h = 1000 * 10 * 20 = 200000 Pa

Suppose air is ideal gas, then if the temperature stays the same, the product of its pressure and volume stays the same

P_1V_1 = P_2V_2

Where P1 = 1.105 Pa is the atmospheric pressure, V_1 is the air volume in the bucket on the suface:

V_1 = Ah

As the pressure increases, the air inside the bucket shrinks. But the crossection area stays constant, so only h, the height of air, decreases:

P_1Ah_1 = P_2Ah_2

h_2 = h_1\frac{P_1}{P_2} = 0.36\frac{1.105}{200000} = 0.000002 m

b) If the temperatures changes, we can still reuse the ideal gas equation above:

\frac{P_1Ah_1}{T_1} = \frac{P_2Ah_2}{T_2}

h_2 = h_1\frac{P_1T_2}{P_2T_1} = 0.36\frac{1.105 * 275}{200000*300} =0.00000182 m

3 0
2 years ago
A 2.70 kg cat is sitting on a windowsill. The cat is sleeping peacefully until a dog barks at him. Startled, the cat falls from
Alchen [17]

Answer:

The speed of the cat when it hits the ground is approximately 7.586 meters per second.

Explanation:

By Principle of Energy Conservation and Work-Energy Theorem, we have that initial potential gravitational energy of the cat (U_{g}), in joules, is equal to the sum of the final translational kinetic energy (K), in joules, and work losses due to air resistance (W_{l}), in joules:

U_{g} = K +W_{l} (1)

By definition of potential gravitational energy, translational kinetic energy and work, we expand the equation presented above:

m \cdot g\cdot h = \frac{1}{2}\cdot m \cdot v^{2}+W_{l} (2)

Where:

m - Mass of the cat, in kilograms.

g - Gravitational acceleration, in meters per square second.

h - Initial height of the cat, in meters.

v - Final speed of the cat, in meters per second.

If we know that m = 2.70\,kg, g = 9.807\,\frac{m}{s^{2}}, h = 5.20\,m and W_{l} = 120\,J, then the final speed of the cat is:

v = \sqrt{\frac{2\cdot (m\cdot g\cdot h-W_{l})}{m} }

v = \sqrt{2\cdot g\cdot h-\frac{W_{l}}{m} }

v \approx 7.586\,\frac{m}{s}

The speed of the cat when it hits the ground is approximately 7.586 meters per second.

4 0
2 years ago
A small mass m is tied to a string of length L and is whirled in vertical circular motion. The speed of the mass v is such that
adell [148]

Answer:

(mv^2/R)/(mg)=1/2

v^2=R/2g

7 0
2 years ago
One of the main factors driving improvements in the cost and complexity of integrated circuits (ICs) is improvements in photolit
nika2105 [10]

Answer:

0.000003782 m

0.000001891 m

0.000001197125 m

Explanation:

\lambda = Wavelength = 248 nm

D = Diameter of beam = 1 cm

f = Focal length = 0.625 cm

The angle is given by

\theta=\dfrac{1.22\lambda}{D}

The width is given by

d=2\theta f\\\Rightarrow d=2\dfrac{1.22\lambda f}{D}\\\Rightarrow d=2\dfrac{1.22\times 248\times 10^{-9}\times 6.25\times 10^{-2}}{1\times 10^{-2}}\\\Rightarrow d=0.000003782\ m

The required width is 0.000003782 m

Minimum resolvable line separation is given by

\dfrac{0.000003782}{2}=0.000001891\ m

The minimum resolvable line separation between adjacent lines is 0.000001891 m

when \lambda=157\ nm

d=2\dfrac{1.22\times 157\times 10^{-9}\times 6.25\times 10^{-2}}{1\times 10^{-2}}\\\Rightarrow d=0.00000239425\ m

The new minimum resolvable line separation between adjacent lines is

\dfrac{0.00000239425}{2}=0.000001197125\ m

6 0
2 years ago
What is the internal energy (to the nearest joule) of 10 moles of Oxygen at 100 K?
kkurt [141]

Answer:

U = 12,205.5 J

Explanation:

In order to calculate the internal energy of an ideal gas, you take into account the following formula:

U=\frac{3}{2}nRT        (1)

U: internal energy

R: ideal gas constant = 8.135 J(mol.K)

n: number of moles = 10 mol

T: temperature of the gas = 100K

You replace the values of the parameters in the equation (1):

U=\frac{3}{2}(10mol)(8.135\frac{J}{mol.K})(100K)=12,205.5J

The total internal energy of 10 mol of Oxygen at 100K is 12,205.5 J

6 0
2 years ago
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