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1 year ago

The graph shows a car's velocity over time. During which time period does the car have the greatest acceleration?

Physics

2 answers:

Oksana_A [137]1 year ago

5 0

Answer:

Between 0 and 1 seconds (B)

Explanation:

The velocity of the car over time is represented by the line graphed here

the steeper the line, the greater change in velocity that occurred in a given time frame.

The steepest portion of the line is between 0-1 seconds, which means that the greatest rate of change occurred between 0-1 seconds.

(acceleration is the rate of change)

vampirchik [111]1 year ago

3 0

C. Between zero and one seconds

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Two skiers arrive at the bottom of a ramp travelling at the same speed. Friction may be ignored in this problem. One skier is an

inysia [295]

Answer:

Both will reach to same height

Explanation:

Here we can see that friction is to be ignored

so we can say that work done by all the non conservative forces is change in mechanical energy

Since all non conservative forces here is zero

so mechanical energy is conserved here

so here we can say that sum of initial kinetic energy and potential energy = sum of final kinetic energy and potential energy

So we will have

$\frac{1}{2}mv^2 = mgH$

now maximum height is given as

$H = \frac{v^2}{2g}$

so here we can say that greatest height will be independent of the mass so they both will reach at same height

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3 years ago

Two charges, one of 2.50μC and the other of -3.50μC, are placed on the x-axis, one at the origin and the other at x = 0.600 m

aev [14]

Answer:

Explanation:

Given

charge of first body $q_1=2.5\ mu C$

charge of second body $q_2=-3.5\ mu C$

Particle 1 is at origin and particle 2 is at $x=0.6\ m$

third Particle which charge +q must be placed left of $2.5\mu C$ because it will repel the q charge while $-3.5\mu C$ will attract it

suppose it is placed at a distance of x m

$F_{1q}=\frac{kq(2.5)}{x^2}$

$F_{2q}=\frac{kq(-3.5)}{(0.6+x)^2}$

$F_{1q}+F_{2q}=0$

$\frac{kq(2.5)}{x^2}+\frac{kq(-3.5)}{(0.6+x)^2}=0$

$\frac{kq(2.5)}{x^2}=\frac{kq(3.5)}{(0.6+x)^2}$

$\frac{0.6+x}{x}=(\frac{3.5}{2.5})^{0.5}$

$0.6+x=1.1832x$

$x=3.27\ m$

5 0

3 years ago

Using diagram differentiate between solenoid and a toroid

damaskus [11]

The Toroid is form when you have wound conductor around circular body. In this case you have magnatic field inside the core but you dont have any poles because circular body dont have ends. This can be used where you want minimum flux leakage and dont need magnatic poles. i.e. toroidal inductor, toroidal transformer.

The Solenoid is forn when you wound conductor around body with limb. In this case magnatic field creates two poles N and S. Solenoids have little bit flux leakage. This used where you want magnatic poles and flux leakage is not an issue. i.e. relay, motors, electromagnates.

1 == toroid

2= solenoid

3 0

3 years ago

Use Newton’s Universal Law of Gravitation to calculate the magnitude of the gravitational force between a 200 kg refrigerator an

expeople1 [14]

Answer:

3.735×10⁻⁶ N

Explanation:

From newton' s law of universal gravitation,

F = Gmm'/r² .............................. Equation 1

Where F = Gravitational force between the person and the refrigerator, m = mass of the person, m' = mass of the refrigerator, r = distance between the person and the refrigerator. G = gravitational universal constant.

Given: m = 70 kg, m' = 200 kg, r = 0.5 m

Constant: G = 6.67×10⁻¹¹ Nm²/kg².

F = (6.67×10⁻¹¹×70×200)/0.5²

F = 93380×10⁻¹¹/0.25

F = 373520×10⁻¹¹

F = 3.735×10⁻⁶ N

Hence the force between the person and the refrigerator = 3.735×10⁻⁶ N

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3 years ago

(a) An elevator of mass m moving upward has two forces acting on it: the upward force of tension in the cable and the downward f

Katen [24]

Answer: T is greater

Explanation:

Since the elevator is moving against gravity more work will be done on the rope

T= m(g+a)

8 0

2 years ago

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