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Pepsi [2]
4 years ago
10

Suppose the roller coaster in Fig. 6-38 (h1 = 33 m, h2 = 13 m, h3 = 25) passes point A with a speed of 2.10 m/s. If the average

force of friction is equal to one fifth of its weight, with what speed will it reach point B? The distance traveled is 35.0 m

Physics
1 answer:
Kamila [148]4 years ago
7 0

Answer:

22.67 m/s

Explanation:

See in the picture.

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As part of your daily workout, you lie on your back and push with your feet against a platform attached to two stiff ideal sprin
shepuryov [24]

Answer:

The magnitude of force must you apply to hold the platform in this position = 888.89 N

Explanation:

Given that :

Workdone (W) = 80.0 J

length x = 0.180 m

The equation  for this  work done by the spring is expressed as:

W = \frac{1}{2}k_{eq}x^2

Making the spring constant k_{eq} the subject of the formula; we have:

k_{eq} = \frac{2W}{x^2}

Substituting our given values, we have:

k_{eq} = \frac{2*80}{0.180^2}

k_{eq} = 4938.27 \ N.m^{-1}

The magnitude of the force that must be apply  to the hold platform in this position is given by the formula :

F = k_{eq}x

F = 4938.27*0.180

F = 888.89 N

6 0
3 years ago
Hydrogen is a nonmetal with many nonmetal properties. Why is it on the periodic table with the metals in Group 1?
mrs_skeptik [129]

Answer:

This is because it has ns1 electron configuration like the alkali metals.

Pls Mark Brainiest, am trying to become Virtuoso

6 0
3 years ago
135g of an unknown substance gains 9133 J of heat as it is heated from 25⁰C to 100⁰C. Using the chart below, determine the ident
telo118 [61]

Answer:

The unknown substance is Aluminum.

Explanation:

We'll begin by calculating the change in the temperature of substance. This can be obtained as follow:

Initial temperature (T₁) = 25 ⁰C

Final temperature (T₂) = 100 ⁰C

Change in temperature (ΔT) =?

ΔT = T₂ – T₁

ΔT = 100 – 25

ΔT = 75 ⁰C

Finally, we shall determine the specific heat capacity of the substance. This can be obtained as follow:

Change in temperature (ΔT) = 75 ⁰C

Mass of the substance (M) = 135 g

Heat (Q) gained = 9133 J

Specific heat capacity (C) of substance =?

Q = MCΔT

9133 = 135 × C × 75

9133 = 10125 × C

Divide both side by 10125

C = 9133 / 10125

C = 0.902 J/gºC

Thus, the specific heat capacity of substance is 0.902 J/gºC

Comparing the specific heat capacity (i.e 0.902 J/gºC) of substance to those given in the table above, we can see clearly that the unknown substance is aluminum.

7 0
3 years ago
125 cm of gas are collected at 15 °C and
aleksklad [387]
The new volume will be ≈26mL
rounded to two significant figures.
Explanation:
This question involves the combined gas law. The equation to use is:
When working with gas laws, the temperature is always in Kelvins. To get Kelvins from a Celsius temperature, add 273.15 to the Celsius temperature.
6 0
3 years ago
0.
larisa [96]

Answer:21

Explanation:

4 0
3 years ago
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