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trapecia [35]
3 years ago
13

In a local bar, a customer slides an empty beer mug down the counter for a refill. The height of the counter is 1.2 m. The mug s

lides off the counter and strikes the floor 1.35 m from the base of the counter. Assume the mug moves in the positive x-direction. Give your answer in three significant figures.
a) With what velocity did the mug leave the counter?
b) What was the direction (in degrees and with respect to the horizontal) of the mug's velocity just before it hit the floor?
Physics
1 answer:
ZanzabumX [31]3 years ago
6 0

Part a can be solve using the equation of trajectory:

Y = x tana + (g*x^2)/ [2(V0^2)*(cos a)^2]

Where y is the height

X is the length

G is the acceleration due to gravity

Vo Is the initail velocity

a is the angle of trajectory

1.2 = 1.35  tan(0) + (9.81*1.35^2)/ [2(V0^2)*(cos 0)^2]

Solve for V0 = 2.729 m/s

b. can be solve using the formula

v = sqrt(2gy)

= sqrt ( 2*1.2*9.81)

= 4.852 m/s  going down ( 0 degree from the horizontal)

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Carbon is allowed to diffuse through a steel plate 15 mm thick. The concentrations of carbon at the two faces are 0.65 and 0.30
beks73 [17]

Answer:

T=575.16K

Explanation:

To solve the problem we proceed to use the 1 law of diffusion of flow,

Here,

J=-D\frac{\Delta C}{\Delta x}

\Delta C is the rate in concentration

\Delta xis the rate in thickness

D is the diffusion coefficient, where,

D= D_0 exp(\frac{Q_d}{RT})

Replacing D in the first law,

J=-(D_0 exp(\frac{-Q_D}{RT}))\frac{\Delta }{\Delta x}

clearing T,

T=\frac{Q_d}{R*ln(\frac{J*\Delta x}{D_0*\Delta C})}

Replacing our values

T=-\frac{80000}{8.31*ln(\frac{(6.2*10^{-7})(-15*10^{-3})}{(1.43*10^{-9})(0.65-0.30)})}

T=-\frac{80000}{-138.09}

T=575.16K

4 0
3 years ago
Visit this website and locate the element mercury, whose chemical symbol is Hg. Click on the square to read about the history, p
Irina-Kira [14]
We simply asked to name three uses for mercury.
The most common and well-known use of mercury is the production of thermometers. It's property to stay liquid at room temperature makes it ideal for a temperature indicator. However, the use of mercury is thermometers has been phased out due to health hazards.
It is also used to form an amalgam which is the result of its combination with silver or gold. Mercury has been used to mine gold and silver. This application has also been phased out.
Today's use of mercury includes mercury-vapor lamps which are the bright lamps used in high-ways.

6 0
2 years ago
Read 2 more answers
Masses A and B rest on very light pistons that enclose a fluid.There is no friction between the pistons and the cylinders they f
RSB [31]

Answer:

D)Not enough information

Explanation:

According to Pascal's principle, the pressure exerted on the two pistons is equal:

p_A = p_B

Pressure is given by the ratio between force F and area A, so we can write

\frac{F_A}{A_A}=\frac{F_B}{A_B}

The force exerted on each piston is just equal to the weight of the corresponding mass: F=W=mg, where m is the mass and g is the gravitational acceleration. So the equation becomes

\frac{m_A g}{A_A}=\frac{m_B g}{A_B}

Now we can rewrite the mass as the product of volume, V, times density, d:

\frac{V_A d_A g}{A_A}=\frac{V_B d_B g}{A_B}

We also know that A_B = 2.0 m^2\\A_A = 1.0 m^2

So we can further re-arrange the equation (and simplify g as well):

\frac{V_A d_A}{1}=\frac{V_B d_B}{2}

\frac{d_A}{d_B}=\frac{V_B}{2V_A}

We are also told that block B has bigger volume than block A: V_B > V_A. However, this information is not enough to allow us to say if the fraction on the right is greater than 1 or smaller than 1: therefore, we cannot conclude anything about the densities of the two objects.

3 0
3 years ago
Skater begins to spend with arms held out at shoulder height. The skater wants to match the speed of the spin to the beat of the
Aleksandr [31]

Answer:

the moment of inertia with the arms extended is Io and when the arms are lowered the moment

I₀/I > 1    ⇒   w > w₀

Explanation:

The angular momentum is conserved if the external torques in the system are zero, this is achieved because the friction with the ice is very small,

           L₀ = L_f

           I₀ w₀ = I w

          w =\frac{I_o}{I} w₀

where we see that the angular velocity changes according to the relation of the angular moments, if we approximate the body as a cylinder with two point charges, weight of the arms

          I₀ = I_cylinder + 2 m r²

where r is the distance from the center of mass of the arms to the axis of rotation, the moment of inertia of the cylinder does not change, therefore changing the distance of the arms changes the moment of inertia.

If we say that the moment of inertia with the arms extended is Io and when the arms are lowered the moment will be

        I <I₀

        I₀/I > 1    ⇒   w > w₀

therefore the angular velocity (rotations) must increase

in this way the skater can adjust his spin speed to the musician.

7 0
3 years ago
Answer these questions about a light year.?
aliina [53]

These are five questions and five answers:

a. State the speed in m/sec using scientific notation

i) Find the place value of the digit with the highest place value (the digit most to the left) in the number 300,000,000: hundred millions.

ii) Determine the number of zeros corresponding to that place value: 9

ii) Write the digit with the highest place value, mulitplied by the power of ten with exponent equal to the number of zeros corresponding to the place of such digit less 1 (9 - 1 = 8): 3 × 10⁸

iii) Complete with the units: 3 × 10⁸ m/s ← answer

b. State the speed in km/sec using scientific notation

i) State the conversion factor: 1 km = 1000 m = 10³ m

⇒ 1 = 1 km / 10³ m

ii) Multiply by the conversion factor:

3 × 10⁸ m/s × 1 km / (10³ m)

iii) Simplify (cancel the units that are common in the numerator and denominator)

3 × (10⁸ / 10³) km/s = 3 × 10⁵ km/s ← answer

c. State the number of seconds in one year using scientific notation

i) Find the place value of the digit with the highest place value (the digit most to the left) in the number 32,000,000: ten millions.

ii) Determine the number of zeros corresponding to that place value: 8

ii) Write the digit with the highest place value, as integer, add the other significant figures as decimals, mulitplied by the power of ten with exponent equal to the number of zeros corresponding to the place of the digit with highest place value less 1 (8 - 1 = 7): 3.2 × 10⁷

iii) Complete with the units: 3.2 × 10⁷ s ← answer

d. Calculate the distance in meters of one light year

i) distance formula: distance = speed × time

ii) Replace with the numbers in scientific notation:

distance = 3 × 10⁸ m/s × 3.2 × 10⁷ s

iii) Simplify (due the operations and cancel the units that are common in the numerator and denominator)

3 × 10⁸ m/s × 3.2 × 10⁷ s = 9.6 × 10 ¹⁵ m ← answer

e. State this distance in centimeters

i) State the conversion factor: 1m = 100 cm = 10² cm

⇒ 1 = 10² cm / 1 m

ii) Multiply by the conversion factor: 9.6 × 10 ¹⁵ m × 10² cm / 1 m

iii) Simplily: 9.6 × 10⁷ cm ← answer


4 0
2 years ago
Read 2 more answers
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