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Monica [59]
3 years ago
6

Which characteristic does an object with a constant acceleration always have?

Physics
2 answers:
Alexus [3.1K]3 years ago
6 0

By definition, speed is the integral of acceleration with respect to time.

We have then:

v = \int\limits^t_0 {a} \, dt

As the acceleration is constant, then integrating we have:

v = a*t + vo

Where,

vo: constant of integration that corresponds to the initial velocity

We observe then that the speed varies linearly when the acceleration is constant .

Therefore, for constant acceleration, the velocity is changing.

Answer:

an object with a constant acceleration always have:

A. changing velocity

VARVARA [1.3K]3 years ago
3 0
The characteristic that an object with constant acceleration will always have is Changing velocity 

Answer. . .
A. C<span>hanging Velocity corresponds with acceleration.

Good luck on your studies, hope this helps~ ^//^</span>
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3 years ago
What is the average speed of particles of atoms at room temperature?
Paha777 [63]

Answer:

300 meters per second. That's equal to about 670 miles per hour.

Explanation:

Not only are air particles incredibly small, they are always moving. And they move fast. At room temperature, they are going about 300 meters per second. That's equal to about 670 miles per hour.

8 0
4 years ago
My Notes A container is divided into two equal compartments by a partition. One compartment is initially filled with helium at a
Ostrovityanka [42]

Answer:

final-temperature = T_{f} = 252.51K

Explanation:

we can solve this problem by using the first law of thermodynamics.

    \Delta U= Q-W

Q= heat added

U= internal energy

W= work done by system

                        E_{final}= E_{initial}

<u> C_{v} (N_{2}) C_{v}(He) T_{f} =C_{v}( N_{2}) T_{N_{2} } C_{v} (He) T_{He}    (1)</u>

C_{v}(N_{2})=1.04\frac{KJ}{Kg K}

C_{v}(He)=5.193\frac{KJ}{Kg K}

now

From equation 1

T_{f}=\frac{C_{v}( N_{2}) T_{N_{2} } C_{v} (He) T_{He}}{C_{v} (N_{2}) C_{v}(He)}

T_{f} = \frac{315\times1.04+5.193\times240}{1.04+5.193}

T_{f} = 252.51K

4 0
4 years ago
Consider the three dip1acement vectors A = (3i - 3j) m, B = (i-4j) m, and C = (-2i + 5j) m. Use the component method to determin
tia_tia [17]

Answer with Explanation:

We are given that

A=3i-3j m

B=i-4 j m

C=-2i+5j m

a.D=A+B+C

D=3i-3j+i-4j-2i+5j

D=2i-2j

Compare with the vector r=xi+yj

We get x=2 and y=-2

Magnitude=\mid D\mid=\sqrt{x^2+y^2}=\sqrt{(2)^2+(-2)^2}=2\sqrt 2 units

By using the formula \mid r\mid=\sqrt{x^2+y^2}

Direction:\theta=tan^{-1}\frac{y}{x}

By using the formula

Direction of D:\theta=tan^{-1}(\frac{-2}{2})=tan^{-1}(-1)=tan^{-1}(-tan45^{\circ})=-45^{\circ}

b.E=-A-B+C

E=-3i+3j-i+4j-2i+5j

E=-6i+12j

\mid E\mid=\sqrt{(-6)^2+(12)^2}=13.4units

Direction of E=\theta=tan^{-1}(\frac{12}{-6}=tan^{-1}(-2)=-63.4^{\circ}

4 0
4 years ago
Please help I'll give brainliest!!
photoshop1234 [79]

Answer:

this may be wrong but when i looked it up it said “Seafloor Mapper”

Explanation:

4 0
4 years ago
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