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Valentin [98]
2 years ago
15

If runner A is running at 7.50 m/s and runner B is running at 7.90 m/s, how long will it take runner B to catch runner A if runn

er A has a
55.0-m head start?
Physics
1 answer:
yaroslaw [1]2 years ago
3 0

Answer:

t= 137.5 s

Explanation:

So if we are wanting to figure out how long it takes runner B to catch runner A. we must first set the slope of each runner equal to one another

<u>Slopes:</u>

Runner A:    y = 7.50x + 55

Runner B:    y = 7.90 x

sooooo

  7.50 x + 55 = 7.90 x

- 7.50 x          - 7.50 x

 55 = .40 x

55/.40 = .40 x / .40

x = 137.5 s

t= 137.5 s

7.50 * 137.5 + 55 =1086.25 m

7.90 * 137.5 =  1086.25 m

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What is total resistor formula
sp2606 [1]

Answer:

If you know the current and voltage across the whole circuit, you can find total resistance using Ohm's Law: R = V / I.

Explanation:

8 0
3 years ago
Witch TWO types of rock are formed as particles get pushed closer together A. Intrusive igneous B. Extrusive igneous C. Sediment
motikmotik

Answer:

C

Explanation:

Sedimentary rocks are the product of

1) weathering of preexisting rocks,

2) transport of the weathering products,

3) deposition of the material

4) compaction (two rocks pushed together)

5) cementation of the sediment to form a rock.

6 0
3 years ago
Read 2 more answers
On his way off to college, Russell drags his suitcase 19 m from the door of his house to the car at a constant speed with a hori
Mashcka [7]

Answer:

The work done on the suitcase is, W = 1691 J

Explanation:

Given data,

The force on the suitcase is, F  = 89 N

The distance Russell dragged the suitcase, S = 19 m

The work done on the suitcase by Russell is equal to the work done on the suitcase to overcome the friction

The work done on the suitcase by Russell is given by the formula

                          W = F · S

Substituting the given values,

                           W = 89 N x 19 m

                           W = 1691 J

Hence, the work done on the suitcase is, W = 1691 J

8 0
4 years ago
Calculate the volume of this regular solid.
Nataly [62]

Answer:

V = 42.41cm^3

Explanation:

In order to calculate the volume of the solid, you use the following formula:

V=\frac{1}{3}\pi r^2 h

where

r: radius of the circular base of the cone = 3 cm

h: height from the circular base to the peak of the cone = 4.5 cm

You replace the values of r and h in the formula for the volume V:

V=\frac{1}{3}\pi(3cm)^2(4.5)=42.411cm^3\approx42.41cm^3

hence, the volume of the solid is 42.41 cm^3

5 0
3 years ago
Review. (c) Assume the dipole is a compass needle-a light bar magnet-with a magnetic moment of magnitude μ . It has moment of in
hammer [34]

The frequency of oscillation is \frac{1}{2\pi }  \sqrt{\frac{\mu B}{I} }.

<h3>What is a magnetic moment?</h3>

The magnetic moment is the magnetism of a magnet or other item that creates a magnetic field, as well as its orientation and strength. Electromagnets, permanent magnets, elementary particles like electrons, different compounds, and a variety of celestial objects are examples of things that have magnetic moments (such as many planets, some moons, stars, etc). The phrase "magnetic moment" typically refers to a system's magnetic dipole moment, which can be represented by an analogous magnetic dipole, which has a magnetic north and south pole that are barely separated from one another. For sufficiently small magnets or at sufficiently wide distances, the magnetic dipole component is adequate. For extended objects, additional terms may be required in addition to the dipole moment, such as the magnetic quadrupole moment.

T = \frac{ Id^{2}\theta }{dt^{2} }

-\mu B\theta=\frac{ Id^{2}\theta }{dt^{2} }

\frac{d^{2}\theta }{dt^{2} } = -(\frac{\mu BI}{\theta} )

By Comparing the above equation with the SHM equation

\frac{d^2 \theta} {dt^{2} } = -\omega^{2} \theta

\omega^{2} =\frac{ \mu B}{I}

Frequency = \frac{\mu}{2\pi }

=\frac{\sqrt{\frac{\mu B}{I} } }{2\pi}

=\frac{1}{2\pi }  \sqrt{\frac{\mu B}{I} }

To learn more about a magnetic moment, visit:

brainly.com/question/17000031

#SPJ4

8 0
1 year ago
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