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Helen
2 months ago
9

At time t=0 , a cart is at x=10 m and has a velocity of 3 m/s in the −x -direction. The cart has a constant acceleration in the

+x -direction with magnitude 3 m/s^2 < a < 6  m/s^2 . Which of the following gives the possible range of the position of the cart at t=1 s ?
Physics
1 answer:
Softa [3K]2 months ago
7 0

Answer:

Explanation:

The lowest acceleration magnitude is 3 m/s².

displacement at t = 1

s = ut + 1 /2 at²

= -3 x 1 + 1/2 x 3 x 1²

= -3 + 1.5

= -1.5 m

position at t = 1 s

= 10 - 1.5

= 8.5 m

The highest acceleration magnitude is 6 m/s².

displacement at t = 1

s = ut + 1 /2 at²

= -3 x 1 + 1/2 x 6 x 1²

= -3 + 3

=  0

position at t = 1 s

= 10 + 0

= 10 m

Thus, the position range is from 8.5 m to 10 m.

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Which graph represents the motion of an object traveling with a positive velocity and a negative acceleration?
Yuliya22 [3333]
Answer
graph 2

Explanation
Velocity refers to the speed at which displacement changes.
Velocity = (change in displacement)/(change in time)
In graph 2, the slope of the curve is positive, indicating a positive velocity.
Acceleration indicates how velocity varies.
The incline of the same curve (graph 2) is diminishing. This suggests that the velocity is falling, signifying negative acceleration.


5 0
1 month ago
Read 2 more answers
Which, if any, of the following statements concerning the work done by a conservative force is NOT true? All of these statements
Yuliya22 [3333]

Answer:

If the starting and ending points are identical, the overall work equals zero.

Explanation:

Option (D) is correct.

A force is considered conservative when the work performed by it while moving an object from point A to point B does not rely on the path taken and remains consistent across all paths. The work done is determined solely by the initial and final locations of the particle. Thus, when the initial and final positions in a conservative field coincide, the work is said to be zero.

8 0
2 months ago
A positive point charge q is placed at the center of an uncharged metal sphere insulated from the ground. The outside of the sph
Softa [3030]
B. The charge on A is -q; B has no charge. Given that a positive charge is situated at the center of an uncharged metallic sphere which is insulated and disconnected from the ground, a negative charge (-q) will appear on the inner surface A of the sphere. Should the exterior surface B be grounded, it will become neutral, resulting in no charge remaining on surface B.
4 0
1 month ago
The deflection plates in an oscilloscope are 10 cm by 2 cm with a gap distance of 1 mm. A 100 volt potential difference is sudde
Maru [3345]

Explanation:

Data provided:

Area A = 10 cm×2 cm = 20×10⁻⁴ m²

Separation distance d between the plates = 1 mm = 1×10⁻³ m

Battery voltage, or emf = 100 V

Resistance = 1025 ohm

Solution:

In an RC circuit, the voltage across the plates varies with time t. At the outset, the voltage matches that of the battery, V₀ = emf = 100V. However, after a certain time t, both the resistance and capacitance alter this, leading to a final voltage V expressed as

V = V_{0}(1-e^{\frac{-t}{RC} } )\\\frac{V}{V_{0} } = 1-e(^{\frac{-t}{RC} }) \\e^{\frac{-t}{RC} } = 1- \frac{V}{V_{0} }

Applying the natural logarithm to both sides,

e^{\frac{-t}{RC} } = 1- \frac{V}{V_{0} } \\\frac{-t}{RC} = ln(1-\frac{V}{V_{0} } )\\t = -RCln(1 - \frac{V}{V_{0} })

t = -RC ln (1-\frac{V}{V_{0} })        (1)

Next, we can determine the capacitance using the plates' area.

C = ε₀A/d

  = \frac{(8.85*10^{-12))} (20*10^{-4}) }{1*10^{-3} }

  = 18×10⁻¹²F

We can now find the time it takes for the voltage to drop from 100 to 55 V by substituting C, V₀, V, and R values into equation (1)

t = -RC ln (1-\frac{V}{V_{0} })

 = -(1025Ω)(18×10⁻¹² F) ln( 1 - 55/100)

 = 15×10⁻⁹s

= 15 ns

5 0
23 days ago
In an isolated system, the total heat given off by warmer substances equals the total heat energy gained by cooler substances. N
Keith_Richards [3271]

Answer:

The temperature of the cooler object was nearly at room temperature. As a result, the system underwent minimal change

Explanation:

In a closed system with two objects at varying temperatures, heat energy typically flows from the hotter object to the cooler one. This transfer is more pronounced when there is a significant temperature disparity between the objects. Conversely, if the temperature difference is minor or negligible, the resulting change will be minimal.

3 0
1 month ago
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