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2 years ago

Find the object's speeds v1, v2, and v3 at times t1=2.0s, t2=4.0s, and t3=13s.

Physics

2 answers:

Burka [1]2 years ago

7 0

Since this is a distance/time graph, the speed at any time is the slope
of the part of the graph that's directly over that time on the x-axis.

At time t1 = 2.0 s
That's in the middle of the first segment of the graph,
that extends from zero to 3 seconds.
Its slope is 7/3 . v1 = 7/3 m/s .

At time t2 = 4.0 s
That's in the middle of the horizontal part of the graph
that runs from 3 to 6 seconds.
Its slope is zero.
v2 = zero .

At time t3 = 13 s.
That's in the middle of the part of the graph that's sloping down,
between 11 and 16 seconds.
Its slope is -3/5 . v3 = -0.6 m/s .

wel2 years ago

7 0

The slope or gradient of a line is a number that describes both the direction and the steepness of the line. For a distance-time graph, it signifies the velocity.

at t1=2.0s
v = 7/3 m/s

At time t2 = 4.0 s
v = zero

At time t3 = 13 s
v = -0.6 m/s

Hope this answers the question. Have a nice day.

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Tex, an 85.0 kilogram rodeo bull rider is thrown from the bull after a short ride. The 520. kilogram bull chases after Tex at 13

Julli [10]

The question above can be answered through using the concept of Conservation of Momentum which may be expressed as,
m1v1 + m2v2 = mTvT
where m1 and v1 are mass and initial velocity of Tex, 2s are that of the bull, and the Ts are the total. Then substituting,
(85 kg)(3 m/s) + (520 kg)(13 m/s) = (520 + 85)(vT)
The value of vT obtained from above equation is 11.6 m/s

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2 years ago

: Two containers have a substantial amount of the air evacuated out of them so that the pressure inside is half the pressure at

ser-zykov [4K]

Complete Question

Two containers have a substantial amount of the air evacuated out of them so that the pressure inside is half the pressure at sea level. One container is in Denver at an altitude of about 6,000 ft and the other is in New Orleans (at sea level). The surface area of the container lid is A=0.0155 m. The air pressure in Denver is PD = 79000 Pa. and in New Orleans is PNo = 100250 Pa. Assume the lid is weightless.

Part (a) Write an expression for the force FNo required to remove the container lid in New Orleans.

Part (b) Calculate the force FNo required to lift off the container lid in New Orleans, in newtons.

Part (c) Calculate the force Fp required to lift off the container lid in Denver, in newtons.

Part (d) is more force required to lift the lid in Denver (higher altitude, lower pressure) or New Orleans (lower altitude, higher pressure)?

Answer:

The expression is $F_{No} = A [P_{No} - \frac{P_{sea}}{2}]$

$F_{No}= 7771.125 \ N$

$F_p = 2.2*10^{6} N$

From the value obtained we can say the that the force required to open the lid is higher at Denver

Explanation:

The altitude of container in Denver is $d_D = 6000 \ ft = 6000 * 0.3048 = 1828.8m$

The surface area of the container lid is $A = 0.0155m^2$

The altitude of container in New Orleans is sea-level

The air pressure in Denver is $P_D = 79000 \ Pa$

The air pressure in new Orleans is $P_{ro} = 100250 \ Pa$

Generally force is mathematically represented as

$F_{No} = \Delta P A$

So we are told the pressure inside is is half the pressure the at sea level so the the pressure acting on the container would

The pressure at sea level is a constant with a value of

$P_{sea} = 101000 Pa$

So the $\Delta P$ which is the difference in pressure within and outside the container is

$\Delta P = P_{No} - \frac{P_{sea}}{2}$

Therefore

$F_{No} = A [P_{No} - \frac{P_{sea}}{2}]$

Now substituting values

$F_{No} = 0.0155 [100250 - \frac{101000}{2}]$

$F_{No}= 7771.125 \ N$

The force to remove the lid in Denver is

$F_p = \Delta P_d A$

So we are told the pressure inside is is half the pressure the at sea level so the the pressure acting on the container would

The pressure at sea level is a constant with a value of

$P_{sea} = 101000 Pa$

At sea level the air pressure in Denver is mathematically represented as

$P_D = \rho g h$

=> $g = \frac{P_D}{\rho h}$

Let height at sea level is h = 1

The air pressure at height $d_D$

$P_d__{D}} = \rho gd_D$

=> $g = \frac{P_d_D}{\rho d_D}$

Equating the both

$\frac{P_D}{\rho h} = \frac{P_d_D}{\rho d_D}$

$P_d_D = P_D * d_D$

Substituting value

$P_d__{D}} = 1828.2 * 79000$

$P_d__{D}} = 1.445*10^{8} Pa$

$\Delta P_d = P_{d} _D - \frac{P_{sea}}{2}$

=> $\Delta P_d = 1.445 *10^{8} - \frac{101000}{2}$

$\Delta P_d = 1.44*10^{8}Pa$

$F_p = \Delta P_d A$

$= 1.44*10^8 * 0.0155$

$F_p = 2.2*10^{6} N$

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3 years ago

Cindy exerts a force of 40 newtons and moves a chair 6 meters. Her brother Andy pushes a different chair for 6 meters while exer

sesenic [268]

Work formula:
W = F * d
F 1 = 40 N, d 1 = 6 m;
F 2 = 30 N; d 2 = 6 m.
W ( Cindy ) = 40 * 6 = 240 Nm
W ( Andy ) = 30 * 6 = 180 Nm
The difference of their amounts if work:
240 Nm - 180 Nm = 60 nm

hope it helps!

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2 years ago

At 5000-kg freight car runs into 10000-kg freight car at rest. they couple upon collision and move with a speed of 2 m/s. what w

aliina [53]

Solution for the problem is:

Total momentum before collision is always equal to total momentum after collision. So note that:
Momentum of car A = 5000 x Xm/s
Momentum of car A + B = 15,000 x 2m/s

So combining the two, will give us the equation:
15,000/5,000 = 3
3 x 2 =6m/s

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2 years ago

A badger is trying to cross the street. Its velocity vvv as a function of time ttt is given in the graph below where rightwards

Mrrafil [7]

Answer: -2.5

Explanation:

1/2(-5)= -2.5

-2.5(1)= -2.5

Got it right in Khan Academy. You’re welcome.

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2 years ago