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aleksklad [387]
3 years ago
10

Which are examples of projectile motion?

Physics
2 answers:
Svetach [21]3 years ago
7 0

Answer:

C and D

Explanation:

an object that is projected at an angle where gravity is the only force acting on the object

ivolga24 [154]3 years ago
7 0

There always are when it says "check all that apply".

For this one, C and D are the ones that apply.

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A wave with a frequency of 14 Hz has a wavelength of 7m. At what speed will this wave travel?
frutty [35]

Answer: A wave with a frequency of 14 Hz has a wavelength of 3 meters. At what speed will this wave travel? 1. = 3m (4. = 42m. 2. ... 1,7m (46) = 7802 m. 4. A wave traveling at 230 m/sec has a wavelength of 2.1 meters. What is the frequency of.

Explanation: please give me brainlest

8 0
3 years ago
Inserting the formulas you found for Xman(t) and Xbus(t) into the conditionXman(tcatch)=Xbus(tcatch) , you obtain the following-
lara [203]

Answer:

c > √(2ab)

Explanation:

In this exercise we are asked to find the condition for c in such a way that the results have been real

The given equation is

              ½ a t² - c t + b = 0

we can see that this is a quadratic equation whose solution is

             t = [c ±√(c² - 4 (½ a) b)]  / 2

for the results to be real, the square root must be real, so the radicand must be greater than zero

              c² -2a b > 0

              c > √(2ab)

3 0
2 years ago
Calculate the amount of heat liberated (in kj) from 411 g of mercury when it cools from 88.0°c to 12.0°c.
Katen [24]
You should note that the melting point of mercury is -38.83°C, while the boiling point is at 356.7°C. Then, that means that there is no latent heat involved here. We only compute for the sensible heat.

ΔH = mCpΔT
The Cp of mercury is 0.14 J/g·°C
Thus,
ΔH = (411 g)(0.14 J/g·°C)(88 - 12°C)
<em>ΔH = 4,373.04 J</em>
5 0
3 years ago
A truck is traveling at 80 m/s and has 12,000 J of kinetic energy. What is the truck’s mass?
Murrr4er [49]

Answer:

3.75 kg

Explanation:

m=2xKE / v²

8 0
2 years ago
A glass flask whose volume is 1000 cm^3 at a temperature of 1.00°C is completely filled with mercury at the same temperature. W
Marat540 [252]

Answer:

the coefficient of volume expansion of the glass is \mathbf{  ( \beta_{glass} )= 1.333 *10^{-5} / K}

Explanation:

Given that:

Initial volume of the glass flask = 1000 cm³ = 10⁻³ m³

temperature of the glass flask and mercury= 1.00° C

After heat is applied ; the final temperature = 52.00° C

Temperature change ΔT = 52.00° C - 1.00° C = 51.00° C

Volume of the mercury overflow = 8.50 cm^3 = 8.50 ×  10⁻⁶ m³

the coefficient of volume expansion of mercury is 1.80 × 10⁻⁴ / K

The increase in the volume of the mercury =  10⁻³ m³ ×  51.00 × 1.80 × 10⁻⁴

The increase in the volume of the mercury = 9.18*10^{-6} \ m^3

Increase in volume of the glass =  10⁻³ × 51.00 × \beta _{glass}

Now; the mercury overflow = Increase in volume of the mercury - increase in the volume of the flask

the mercury overflow = (9.18*10^{-6}  -  51.00* \beta_{glass}*10^{-3})\ m^3

8.50*10^{-6} = (9.18*10^{-6}  -51.00* \beta_{glass}* 10^{-3} )\ m^3

8.50*10^{-6} - 9.18*10^{-6} = ( -51.00* \beta_{glass}* 10^{-3} )\ m^3

-6.8*10^{-7} =  ( -51.00* \beta_{glass}* 10^{-3} )\ m^3

6.8*10^{-7} =  ( 51.00* \beta_{glass}* 10^{-3} )\ m^3

\dfrac{6.8*10^{-7}}{51.00 * 10^{-3}}=  ( \beta_{glass} )

\mathbf{  ( \beta_{glass} )= 1.333 *10^{-5} / K}

Thus; the coefficient of volume expansion of the glass is \mathbf{  ( \beta_{glass} )= 1.333 *10^{-5} / K}

5 0
2 years ago
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