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Dominik [7]
3 years ago
9

Daniel rode his bike 35km in 6 hours. What was his speed?

Physics
2 answers:
spayn [35]3 years ago
8 0
His speed per hour was 5.8km and the total of 6 hours is 35km
Ilya [14]3 years ago
8 0
Daniel was riding his bike at 5.83km/hr
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A piston raises a weight, then lowers it again to its original height. If the process the system follows as the piston rises is
FinnZ [79.3K]

Answer:

Negative

Explanation:

First law of thermodynamic also known as the  law of conservation of energy states that the total energy of an isolated system is constant; energy can be transformed from one form to another, but can be neither created nor destroyed.

The first law relates  relates changes in internal energy to heat added to a system and the work done by a system by the conservation of energy.

The first law is mathematically given as ΔU  = U_{F} - U_{0} = Q - W

Where Q  = Quantity of heat

            W = Work done

From the first law The internal energy has the symbol U. Q is positive if heat is added to the system, and negative if heat is removed; W is positive if work is done by the system, and negative if work is done on the system.

Analyzing the pistol when it raises in isothermal and when it falls in  isobaric state.The following can be said:

In the Isothermal compression of a gas there is work done on the system to decrease the volume and increase the pressure. For work to be done on the system it is a negative work done then.

In the Isobaric State An isobaric process occurs at constant pressure. Since the pressure is constant, the force exerted is constant and the work done is given as PΔV.If a gas is to expand at a constant pressure, heat should be transferred into the system at a certain rate.Isobaric is a fuction of heat which is Isothermal Provided the pressure is kept constant.

In Isobaric definition above it can be seen that " Heat should be transferred into the system ata certain rate. For heat to be transferred into the system work is deinitely been done on the system thereby favouring the negative work done.

6 0
3 years ago
The components of a 15 meters per second velocity at an angle of 60 degrees above the horizontal are?
mafiozo [28]

Answer:

x-component of velocity: 7.5 m/s

y-component of velocity: 13 m/s

Explanation:

This problem is pure trigonometry. Assuming you know trig, there are only a couple of steps to solving this problem. First, split the velocity into components; recall that any vector not directed along an axis has x and y components. Then, remember that sinΘ = opposite/hypotenuse. Applying this to your scenario, you get sin60° = vy/15. Multiplying this out gives you vy=15sin60. Put this into a calculator (make sure it's set to degree mode because the angle in this problem is in degrees) and you should get 12.99, which you can round up to 13 m/s. This is the velocity in the y-direction.

The procedure to find the x-velocity is very similar, but instead of using sine, we will use the cosine of theta. Recall that cosΘ=adjacent/hypotenuse. Once again plugging this scenario's numbers into that, you end up with cos60 = vₓ/15. Multiplying this out gives you vₓ = 15cos60. Once again, plug this into your calculator. 7.5 m/s should be your answer. This is the velocity in the x-direction.

By the way, a quick way to find the components of a vector, whether it's velocity, force, or whatever else, is to use these functions. Generally, if the vector points somewhere that's not along an axis, you can use this rule. The x-component of the vector is equal to hypotenuse*cosΘ and the y-component of the vector is equal to hypotenuse*sinΘ.

8 0
2 years ago
A ball rolls over the edge of a platform with only a horizontal velocity. The height of the platform is 1.60m and the horizontal
AysviL [449]

Answer:

v = 46.99 m/s

Explanation:

The velocity of the ball just before it touches the ground, is given by the following formula:

v=\sqrt{v_x^2+v_y^2}           (1)

vx: horizontal component of the velocity

vy: vertical component of the velocity

The vertical component vy is calculated by using the following formula:

v_y^2=v_{oy}^2+2gh   (2)

vy: final velocity

voy: initial vertilal velocity = 0m/s  (because it is a semi parabolic motion)

g: gravitational acceleration = 9.8 m/s^2

h: height = 1.60m

You replace the values of the parameters in the equation (2):

v_y=2(9.8m/s^2)(1.60m)=31.36\frac{m}{s}

vx is calculated by using the information about the horizontal range of the ball:

R=v_o\sqrt{\frac{2h}{g}}    (3)

R: horizontal range of the ball = 20.0 m

You solve the previous equation for vo, the initial horizontal velocity:

v_o=R\sqrt{\frac{g}{2h}}=(20.0m)\sqrt{\frac{9.8m/s^2}{2(1.60m)}}\\\\v_o=35\frac{m}{s}

The horizontal component of the velocity is constant in the complete trajectory, hence, you have that

vx = vo = 35 m/s

Finally, you replace the values of vx and vy in the equation (1):

v=\sqrt{(35m/s)^2+(31.36m/s)^2}=46.99\frac{m}{s}

The velocity of the ball just before it touches the ground is 46.99 m/s

5 0
4 years ago
A 15 kg kangaroo jumps with an upward acceleration of 3 m/s2 by pushing hard off the ground.
patriot [66]

Explanation:

The net force would be upwards since the kangaroo would have to overcome gravity to jump

3 0
3 years ago
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An electron in the first energy level of the electron cloud has an electron in the third energy level
lukranit [14]

Answer:

a lower energy than

Explanation:

3 0
4 years ago
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