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3 years ago

In the graph, which region shows nonuniform positive acceleration?

Physics

2 answers:

cricket20 [7]3 years ago

3 0

Answer: A.AB

Explanation:

This Velocity vs Time graph shows the acceleration of a body or object, since acceleration is the variation of velocity in time.

As we can see in the attached image, the graph can be divided in four segments:

OA: In this segment the acceleration is changing at a uniform rate. In addition we can see it has a positive slope, hence we are dealing with a positive uniform acceleration.

AB: In this segment the acceleration is changing at a nonuniform rate, since in this part it is not possible to calculate the slope. However if this were uniform, the slope woul be positive. This means the <u>acceleration is nonuniform and positive.</u>

BC: In this segment the acceleration is changing at a nonuniform rate, since in this part it is not possible to calculate the slope. However if this were uniform, the slope woul be negative. This means the acceleration is nonuniform and negative.

CD: In this segment the acceleration is changing at a uniform rate. In addition we can see it has a negative slope, hence we are dealing with a negative uniform acceleration.

From all these segments, the only one that fulfils the nonuniform positive acceleration condition is option A:

Segment AB

lesya692 [45]3 years ago

3 0

Answer:

AB plato users ;)

Explanation:

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In your experiment, you measure a total deflection of 4.12 cm when an electric field of 1.10×103V/m is established between the p

Bond [772]

Answer:

$B_0 = 1.69 \times 10^{-4}\ T$

Explanation:

given,

total deflection = 4.12 cm

Electric field = 1.1 ×10³ V/m

plate length = 6 cm

distance between them = 12 cm

using formula

$v_0 = \sqrt{\dfrac{q\epsilon_0d}{ym}(\dfrac{d}{2}+L)}$

q = 1.6 × 10⁻¹⁹ C

m = 9.11 x 10⁻³¹ kg

d = 0.06 m

L = 0.12 m

$v_0 = \sqrt{\dfrac{1.6 \times 10^{-19}\times 1.1 \times 10^{3}\times 0.06}{0.0412\times 9.11 \times 10^{-31} }(\dfrac{0.06}{2}+0.12)}$

v_0 = 6496355.63 m/s

$v_0 = \dfrac{E}{B_0}$

$B_0 = \dfrac{E}{v_0}$

$B_0 = \dfrac{1.1\times 10^{3}}{6496355.63}$

$B_0 = 1.69 \times 10^{-4}\ T$

5 0

3 years ago

A stone is catapulted at time t = 0, with an initial velocity of magnitude 19.9 m/s and at an angle of 39.9° above the horizonta

larisa [96]

Answer:

Part a)

$x = 15.76 m$

Part b)

$y = 7.94 m$

Part c)

$x = 26.16 m$

Part d)

$y = 7.49 m$

Part e)

$x = 83.23 m$

Part f)

$y = -75.6 m$

Explanation:

As we know that catapult is projected with speed 19.9 m/s

so here we have

$v_x = 19.9 cos39.9$

$v_x = 15.3 m/s$

similarly we have

$v_y = 19.9 sin39.9$

$v_y = 12.76 m/s$

Part a)

Horizontal displacement in 1.03 s

$x = v_x t$

$x = (15.3)(1.03)$

$x = 15.76 m$

Part b)

Vertical direction we have

$y = v_y t - \frac{1]{2}gt^2$

$y = (12.76)(1.03) - 4.9(1.03)^2$

$y = 7.94 m$

Part c)

Horizontal displacement in 1.71 s

$x = v_x t$

$x = (15.3)(1.71)$

$x = 26.16 m$

Part d)

Vertical direction we have

$y = v_y t - \frac{1]{2}gt^2$

$y = (12.76)(1.71) - 4.9(1.71)^2$

$y = 7.49 m$

Part e)

Horizontal displacement in 5.44 s

$x = v_x t$

$x = (15.3)(5.44)$

$x = 83.23 m$

Part f)

Vertical direction we have

$y = v_y t - \frac{1]{2}gt^2$

$y = (12.76)(5.44) - 4.9(5.44)^2$

$y = -75.6 m$

6 0

3 years ago

A substance did not change its chemical nature in a reaction. Which most likely describes the reaction?

Andre45 [30]

B. It was hit with a hammer.

4 0

3 years ago

Read 2 more answers

A student is examining a bacterium under the microscope. The E. coli bacterial cell has a mass of m = 0.900 fg (where a femtogra

just olya [345]

Answer:

9.73 x 10⁻¹⁰ m

Explanation:

According to Heisenberg uncertainty principle

Uncertainty in position x uncertainty in momentum ≥ h / 4π

Δ X x Δp ≥ h / 4π

Δp = mΔV

ΔV = Uncertainty in velocity

= 2 x 10⁻⁶ x 3 / 100

= 6 x 10⁻⁸

mass m = 0.9 x 10⁻¹⁵ x 10⁻³ kg

m = 9 x 10⁻¹⁹

Δp = mΔV

= 9 x 10⁻¹⁹ x 6 x 10⁻⁸

= 54 x 10⁻²⁷

Δ X x Δp ≥ h / 4π

Δ X x 54 x 10⁻²⁷ ≥ h / 4π

Δ X = h / 4π x 1 / 54 x 10⁻²⁷

= $\frac{6.6\times10^{-34}}{4\times3.14\times54\times10^{-27}}$

= 9.73 x 10⁻¹⁰ m

7 0

3 years ago

Suppose that you have been chosen for a space mission to a distant planet. Due to the length of time you'll be away from Earth y

Contact [7]

Answer:

I should be active for 15 hours to meet the physical activity requirement.

Explanation:

Since time dilates in moving objects, we use the formula t = t₀/√(1 - β²) where t = time in space vehicle, t₀ = time on earth = 9 hours and β = v/c where v = speed of space vehicle = 0.8c.

So, t = t₀/√(1 - β²)

t = 9/√(1 - (v/c)²)

= 9/√(1 - (0.8c/c)²)

= 9/√(1 - (0.8)²)

= 9/√(1 - (0.64)

= 9/√0.36

= 9/0.6

= 15 hr

So, according to a timer on the space vehicle, I should be active for 15 hours to meet the physical activity requirement.

8 0

3 years ago