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weqwewe [10]
3 years ago
8

A 10-kg object is pushed across a rough surface. It begins at rest and accelerates to a speed of 4 m/s in a distance of 5m. Whic

h fact(s) support the conclusion that the object experiences an unbalanced force?
Physics
1 answer:
brilliants [131]3 years ago
5 0

Answer:

Explanation:

Unbalanced Force according to newton's second law is the one which causes  the object to move or break its state of rest on application of force.

Here the object accelerate to a speed of 4 m/s and moves a distance of 5 m on application of force .

Thus we can say that the applied force is unbalanced in nature.

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Two bicyclists are accelerating forward in a straight line, and Biker 1 has less mass than Biker 2. If the net force on the bike
Solnce55 [7]

If  Biker 1 has less mass than Biker 2 it the follows that biker 1 has a greater acceleration than Biker 2.​

<h3>What is the Newton second law?</h3>

The Newton second law gives the relationship between the force and the acceleration. We know that; acceleration = Force/mass.

Now if Biker 1 has less mass than Biker 2 it the follows that biker 1 has a greater acceleration than Biker 2.​

Learn more about Newton's law:brainly.com/question/27573481

#SPJ1

4 0
2 years ago
A tank contains gas at 13.0°C pressurized to 10.0 atm. The temperature of the gas is increased to 95.0°C, and half the gas is re
fomenos

Answer:

The pressure of the remaining gas in the tank is 6.4 atm.

Explanation:

Given that,

Temperature T = 13+273=286 K

Pressure = 10.0 atm

We need to calculate the pressure of the remaining gas

Using equation of ideal gas

PV=nRT

For a gas

P_{1}V_{1}=nRT_{1}

Where, P = pressure

V = volume

T = temperature

Put the value in the equation

10\times V=nR\times286....(I)

When the temperature of the gas is increased

Then,

P_{2}V_{2}=\dfrac{n}{2}RT_{2}....(II)

Divided equation (I) by equation (II)

\dfrac{P_{1}V}{P_{2}V}=\dfrac{nRT_{1}}{\dfrac{n}{2}RT_{2}}

\dfrac{10\times V}{P_{2}V}=\dfrac{nR\times286}{\dfrac{n}{2}R368}

P_{2}=\dfrac{10\times368}{2\times286}

P_{2}= 6.433\ atm

P_{2}=6.4\ atm

Hence, The pressure of the remaining gas in the tank is 6.4 atm.

4 0
3 years ago
Based on relative bond strengths, classify these reactions as endothermic (energy absorbed) or exothermic (energy released)
cluponka [151]
<span>Exothermic reaction is a chemical reaction that releases energy. This reaction releases heat energy or light . In an endotermic reaction energy is used. Enthaply is the heat energy change , delta H. If the sum of the enthalpies of the reactans is greater than the products the reaction is exothermic. If the products side has a larger enthaply than the process is endothermic. So, if delta H is negative then the process is exothermic. If delta H is positive, than the process is endothermic. Exothermic are: A+BC -> AB+C A2+B2 -> 2AB Endothermic are:AB+C -> AC+B A2 + C2 -> 2AC B2+C2 -> 2BC</span>
5 0
4 years ago
In a darkened room, a burning candle is placed 1.84 m from white wall. A lens is placed between candle and wall at a location th
KATRIN_1 [288]

Answer:

82.4 cm

Explanation:

The object and screen are kept fixed ie the distance between them is fixed and by displacing lens between them images are formed on the screen . In the first case let u be the object distance and v be the image distance

then ,

u + v = 184 cm

In the second case of image formation , v becomes u and u becomes v only then image formation in the second case is possible.

The difference between two object distance ie(  v - u ) is the distance by which lens is moved so

v - u = 82.4 cm

7 0
3 years ago
Many web sites describe how to add wires to your clothing to keep you warm while riding your motorcycle. The wires are added to
Tomtit [17]

Answer:

P=42.075W

Explanation:

The power provided by a resistor (wire in this case) is given by:

P=\frac{V^2}{R}.

The resistance of a wire is given by:

R=\frac{\rho L}{A}

Where for the resistivity the one of the copper should be used: \rho=1.68\times10^{-8}\Omega m.

The area A is that of a circle, which written in terms of its diameter is:

A=\pi r^2=\pi (d/2)^2=\frac{\pi d^2}{4}

Putting all together:

P=\frac{AV^2}{\rho L}=\frac{\pi d^2V^2}{4\rho L}

Which for our values is:

P=\frac{\pi (0.00025m)^2(12V)^2}{4(1.68\times10^{-8}\Omega m)(10m)}=42.075W

7 0
3 years ago
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