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2 years ago

An object starts at 11 m/s with an acceleration of 12 m/s? What is its velocity after 4.5 seconds?

Physics

1 answer:

marshall27 [118]2 years ago

5 0

Answer:

65 m/s

Explanation:

v=v0+at <=> v = 11 + 12 t ∧ t = 4.5 s <=> v = 11 + 12×4.5 <=> v = 65 m/s

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A frictionless cart of mass M is attached to a spring with spring constant k. When the cart is displaced 6 cm from its rest posi

Marianna [84]

Answer:

Time period of horizontal Oscillation = T = 2 $\pi$ $\sqrt{\frac{m}{k} }$

As you can see, from the equation, Period of oscillation depends upon the mass and the spring constant. Not on the displacement.

Explanation:

Solution:

As we know that:

F = Kx

Here,

K = Spring constant

x = displacement.

First, they are displacing it with 6 cm from its rest position for which Time period of the oscillation is T = 2 seconds.

But next, they want to know the effect on the time period of the oscillation if the displacement x is doubled from 6cm to 12 cm.

First of all, let us see the equation of the time period of the oscillation.

We need to check, if time period does depend on the displacement or not.

As we know,

Time period of horizontal Oscillation = T = 2 $\pi$ $\sqrt{\frac{m}{k} }$

As you can see, from the equation, Period of oscillation depends upon the mass and the spring constant. Not on the displacement.

Since, K is the constant for a particular spring, we need to change the mass of the cart to change the time period.

Hence the Time period will remain same.

8 0

2 years ago

Jose’s lab instructor gives him a solution of sodium phosphate that is buffered to a pH of 4. Because of an error that he made w

Kazeer [188]

I am thinking that maybe the problem is not with the calibration. It might be that the buffered solution is already expired since at this point the solution is already not stable and will give a different pH reading than what is expected.

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3 years ago

An insulated pipe carries steam at 300°C. The pipe is made of stainless steel (with k = 15 W/mK), has an inner diameter is 4 cm,

insens350 [35]

Answer:

The answers to the question are

(i) The rate of heat loss per-unit-length (W/m) from the pipe is 131.62 W

(ii) The temperature of the outer surface of the insulation is 49.89 °C

Explanation:

To solve the question, we note that the heat transferred is given by

$Q = \frac{2\pi L(t_{hf} - t_{cf}) }{\frac{1}{h_{hf}r_1}+\frac{ln(r_2/r_1)}{k_A} + \frac{ln(r_3/r_2)}{k_B} +\frac{1}{h_{cf}r_3}}$

Where

$t_{hf}$ = Temperature at the inside of the pipe = 300 °C

$t_{f}$ = Temperature at the outside of the pipe = 20 °C

r₁ =internal radius of pipe = 4.0 cm

r₂ = Outer radius of pipe = 4.5 cm

r₃ = Outer radius of the insulation = r₂ + 2.5 = 7.0 cm

$k_A$ = 15 W/m·K

$k_B$ = 0.038 W/m·K

$h_{hf}$ = 75 W/m²·K

$h_{cf}$ = 10 W/m²·K

Plugging in the values in the above equation where for a unit length L = 1 m, we have

Q = 131.32 W

From which we have, for the film of air at the pipe outer boundary layer

$Q = \frac{t_A-t_B}{R_T}$ Where $R_T$ for the air film on the pipe outer surface is given by

$R_T= \frac{1}{\alpha A}$

where A =area of the outside of the pipe

= $\frac{1}{10*2\pi*0.07*1 }$ = 0.227 K/W

Therefore

131.32 W = $\frac{t_A-20}{0.227}$ which gives

$t_A$ = 49.89 °C

Heat transferred by radiation = q' = ε×σ×(T₁⁴ - T₂⁴)

Where ε = 0.9, σ, = 5.67×10⁻⁸W/m²·(K⁴)

T₁ = Surface temperature of the pipe = 49.89 °C and

T₂ = Temperature of the surrounding = 20.00 °C

Plugging in the values gives, q' = 0.307 W per m²

Total heat lost per unit length = 131.32 + 0.307 =131.62 W

8 0

2 years ago

A washing machine heats 10kg of water in each wash cycle. How much energy is saved by washing at 30'c instead of 50'c if the sta

Volgvan

The equation for this is very simple you add then you subtract then you get the answer then you divide then it all works out for you

6 0

3 years ago

Formula for percentage error

GarryVolchara [31]

Answer:

PE = (|accepted value – experimental value| \ accepted value) x 100%

Explanation:

7 0

2 years ago