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Elodia [21]
3 years ago
12

A speedboat travels from the dock to the first buoy, a distance of 20 meters, in 18 seconds. It began the trip at a speed of o m

/s. By the time it reached the buoy, it was going 10 m/s. What was the average velocity of the boat?
A) -3.33 m/s^2
B) -1.11 m/s
C) +1.11 m/s
D) +3.33 m/s
Physics
2 answers:
kifflom [539]3 years ago
8 0

its +1.11 I TOOK THE TEST

MArishka [77]3 years ago
5 0
If you or any else needs the answer for this it is C. +1.11 m/s.
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If a force of 1.0 N pulled up and parallel to the surface of the incline is required to raise the mass back to the top of the in
wel

The question is incomplete! The complete question along with answer and explanation is provided below.

Question:

A 0.5 kg mass moves 40 centimeters up the incline shown in the figure below. The vertical height of the incline is 7 centimeters.

What is the change in the potential energy (in Joules) of the mass as it goes up the incline?  

If a force of 1.0 N pulled up and parallel to the surface of the incline is required to raise the mass back to the top of the incline, how much work is done by that force?

Given Information:  

Mass = m = 0.5 kg

Horizontal distance = d = 40 cm = 0.4 m

Vertical distance = h = 7 cm = 0.07 m

Normal force = Fn = 1 N

Required Information:  

Potential energy = PE = ?

Work done = W = ?

Answer:

Potential energy = 0.343 Joules

Work done = 0.39 N.m

Explanation:

The potential energy is given by

PE = mgh

where m is the mass of the object, h is the vertical distance and g is the gravitational acceleration.

PE = 0.5*9.8*0.07

PE = 0.343 Joules

As you can see in the attached image

sinθ = opposite/hypotenuse

sinθ = 0.07/0.4

θ = sin⁻¹(0.07/0.4)

θ = 10.078°

The horizontal component of the normal force is given by

Fx = Fncos(θ)

Fx = 1*cos(10.078)

Fx = 0.984 N

Work done is given by

W = Fxd

where d is the horizontal distance

W = 0.984*0.4

W = 0.39 N.m

3 0
3 years ago
2 Polnis
Marta_Voda [28]

Answer:

Oprion A

The length of time over which the conditions are measured

Explanation:

Weather captures the daily atmospheric conditions over a short duration hence it is short-term. Climate is the average of weather and covers a longer duration hence long-term. Therefore, what differentiatea these two is the length of time over which the conditions are measured.

7 0
3 years ago
A mercury thermometer reads 10oC when dipped into melting ice and 90oC
Crank

Answer:

Thermometer will read 26 degrees Celsius.

Please vote for Brainliest and I hope this helps!

3 0
2 years ago
A 21.6−g sample of an alloy at 93.00°C is placed into 50.0 g of water at 22.00°C in an insulated coffee-cup calorimeter with a h
IrinaK [193]

Answer : The specific heat capacity of the alloy 1.422J/g^oC

Explanation :

In this problem we assumed that heat given by the hot body is equal to the heat taken by the cold body.

q_1=-q_2

m_1\times c_1\times (T_f-T_1)=-m_2\times c_2\times (T_f-T_2)

where,

C_1 = specific heat of alloy = ?

C_2 = specific heat of water = 4.18J/g^oC

m_1 = mass of alloy = 21.6 g

m_2 = mass of water = 50.0 g

T_f = final temperature of system = 31.10^oC

T_1 = initial temperature of alloy = 93.00^oC

T_2 = initial temperature of water = 22.00^oC

Now put all the given values in the above formula, we get

21.6g\times c_1\times (31.10-93.00)^oC=-50.0g\times 4.18J/g^oC\times (31.10-22.00)^oC

c_1=1.422J/g^oC

Therefore, the specific heat capacity of the alloy 1.422J/g^oC

6 0
3 years ago
a sample contains 100 g of radioactive isotope. How much radioactive isotope will remain in the sample after 1 half-life?
kap26 [50]

Answer:

\huge\boxed{50g}

Definition:

Half-life- The time taken for half of the radioactive isotopes to decay.

Explanation:

How does radioactive decay work? Radioactive decay is a process by which unstable nuclei become more stable through the emission of alpha or beta particles or gamma rays.

Since a half-life is the time taken for half of the isotopes to decay, we can simply divide the initial mass of 100 grams by 2; this gives us 50 grams.

1) Divide 100g by 2.

\frac{100g}{2}=50g

8 0
2 years ago
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