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3 years ago

The velocity time graph of an object is shown below. How far does the object travel in the time interval t =4 s to t = 6 s?

Physics

1 answer:

Trava [24]3 years ago

3 0

The distance covered by the object between t =4 s and t = 6 s is 4 m

Explanation:

In a velocity-time graph, the distance covered by the object represented can be found by calculating the area under the curve.

Therefore, the distance covered by the object between t = 4 s and t = 6 s is the area under the curve between 4 s and 6 s.

We see that we have to calculate the area of a triangle, with:

Base:

$b=6-4 = 2$

And height:

$h=4-0 = 4$

Therefore, the area is

$A=\frac{1}{2}bh=\frac{1}{2}(2)(4)=4$

So, the distance covered by the object is 4 m.

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A disk of mass M and radius R rotates at angular velocity ω0. Another disk of mass M and radius r is dropped on top of the rotat

AleksandrR [38]

Answer:

$\omega = \frac{(R^2\omega_o}{(R^2 + r^2)}$

Explanation:

As we know that there is no external torque on the system of two disc

then the angular momentum of the system will remains conserved

So we will have

$L_i = L_f$

now we have

$L_i = (\frac{1}{2}MR^2)\omega_o$

also we have

$L_f = (\frac{1}{2}MR^2 + \frac{1}{2}Mr^2)\omega$

now from above equation we have

$(\frac{1}{2}MR^2)\omega_o = (\frac{1}{2}MR^2 + \frac{1}{2}Mr^2)\omega$

now we have

$\omega = \frac{MR^2\omega_o}{(MR^2 + Mr^2)}$

$\omega = \frac{(R^2\omega_o}{(R^2 + r^2)}$

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3 years ago

What is the speed of a bobsled whose distance-time graph indicates that it traveled 113m in 29s?

Elza [17]

In this item, we are asked to determine the speed of the bobsled given the distance traveled and the time it takes to cover the certain distance. This can mathematically be expressed as,
speed = distance / time

Substituting the given values in this item,
speed = (113 m) / (29 s)
speed = 3.90 m/s

<em>ANSWER: 3.90 m/s</em>

3 0

3 years ago

An element's atomic number is 58. How many protons would an atom of this element have?

agasfer [191]

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4 0

3 years ago

Ow much charge flows from a 12.0 v battery when it is connected across a completely discharged 18.0 μf capacitor

Trava [24]

The equation Q=CV (Charge = product of Capacitance and potential difference) tells us that the maximum charge that can be stored on a capacitor is equal to the product of it's capacitance and the potential difference across it. In this case the potential difference across the capacitor will be 12.0V (assuming circuit resistance is negligable) and it has a capacitance of 18.0μf or 18.0x10^-6f, therefore charge equals (18.0x10^-6)x12=2.16x10^-4C (Coulombs).

5 0

3 years ago

A 600-turn solenoid, 25 cm long, has a diameter of 2.5 cm. A 14-turn coil is wound tightly around the center of the solenoid. If

Delvig [45]

Answer:

The induced emf in the short coil during this time is 1.728 x 10⁻⁴ V

Explanation:

The magnetic field at the center of the solenoid is given by;

B = μ(N/L)I

Where;

μ is permeability of free space

N is the number of turn

L is the length of the solenoid

I is the current in the solenoid

The rate of change of the field is given by;

$\frac{\delta B}{\delta t} = \frac{\mu N \frac{\delta i}{\delta t} }{L} \\\\\frac{\delta B}{\delta t} = \frac{4\pi *10^{-7} *600* \frac{5}{0.6} }{0.25}\\\\\frac{\delta B}{\delta t} =0.02514 \ T/s$

The induced emf in the shorter coil is calculated as;

$E = NA\frac{\delta B}{\delta t}$

where;

N is the number of turns in the shorter coil

A is the area of the shorter coil

Area of the shorter coil = πr²

The radius of the coil = 2.5cm / 2 = 1.25 cm = 0.0125 m

Area of the shorter coil = πr² = π(0.0125)² = 0.000491 m²

$E = NA\frac{\delta B}{\delta t}$

E = 14 x 0.000491 x 0.02514

E = 1.728 x 10⁻⁴ V

Therefore, the induced emf in the short coil during this time is 1.728 x 10⁻⁴ V

8 0

3 years ago