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3 years ago

How do i do this???/

Physics

2 answers:

ikadub [295]3 years ago

8 0

If you take the Western direction as the positive direction and the Eastern direction as negative one, then,

Distance traveled to the West = +15 m

Distance traveled to the east = (-2 m) + (-7 m ) = -9 m

Total Displacement = 15 m + (-9) m

Total displacement = 6 m

Since the total displacement is positive, the displacement will be in the Western direction.

So, option (D) is correct.

Gekata [30.6K]3 years ago

5 0

Start at the big tree, and take the instructions one at a time: 1). Walk 15 west. 2). From there, turn around and walk 2 east. 3). From there, walk 7 more east. Where are you from the big tree now ? Aren't you 6 west of it ? I mean, let's see some common sense here !

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Transverse waves are sent along a 4.50 m long string with a speed of 85.00 m/s. The string is under a tension of 20.00 N. What i

frutty [35]

Answer:

m = 0.0125 kg

Explanation:

Let us apply the formula for the speed of a wave on a string that is under tension:

$v = \sqrt{\frac{F}{\mu} }$

where F = tension force

μ = mass per unit length

Mass per unit length is given as:

μ = m / l

where m = mass of the string

l = length of the string

This implies that:

$v = \sqrt{\frac{F}{m/l} }\\ \\v = \sqrt{\frac{F * l}{m} }$

Let us make mass, m, the subject of the formula:

$v^2 = \frac{F * l}{m}\\\\m = \frac{F * l}{v^2}$

From the question:

F = 20 N

l = 4.50 m

v = 85 m/s

Therefore:

$m = \frac{20 * 4.5}{85^2}\\\\m = \frac{90}{7225}\\ \\m = 0.0125 kg$

5 0

2 years ago

A ball having a mass of 0.20 kilograms is placed at a height of 3.25 meters. If it is dropped from this height, what will be the

asambeis [7]

EC_1 + EP_1 = EC2 + EP_2

EC_2 = 0

EC_2 = EP_1 - EP_2

EC_2 = mg(H_1 - H_2) = 0.20 kg * 9.8 m/s^2 * (3.25 m - 1.5m) = 3.43 J

7 0

2 years ago

Read 2 more answers

Two carts have a compressed spring between them and are initially at rest. One of the carts has total mass, including its conten

ladessa [460]

Answer:

A) - 1.8 m/s

Explanation:

As we know that whole system is initially at rest and there is no external force on this system

So total momentum of the system must be conserved

so we will have

$m_1v_1 + m_2v_2 = 0$

now plug in all data into above equation

$5(v) + 3(3)$

$5v = -9$

$v = -1.8 m/s$

so correct answer is

A) - 1.8 m/s

3 0

3 years ago

. Consider a fully extended arm that is rotating about the shoulder such that with a shoulder-to-hand length of 30 cm. If the ar

Rainbow [258]

Answer:

13.309 m/s²

Explanation:

Length from shoulder to hand, l = 30 cm = 0.3 m

initial velocity, u = 1 m/s

final velocity, v = 2.5 m/s

time, t = 3 s

Let the tangential acceleration is a.

by using first equation of motion

v = u + at

2.5 = 1 + 3 a

a = 0.5 m/s²

Let the centripetal acceleration is a'.

a' = v'²/l

a' = 2 x 2 / 0.3

a' = 13.3 m/s²

The tangential acceleration and the centripetal acceleration are both perpendicular to each other. So, the net acceleration is given by

$A=\sqrt{a^{2}+a'^{2}}$

$A=\sqrt{0.5^{2}+13.3^{2}}$

A = 13.309 m/s²

3 0

3 years ago

Radar uses radio waves of a wavelength of 2.4 ${\rm m}$ . The time interval for one radiation pulse is 100 times larger than t

blondinia [14]

Answer:

120 m

Explanation:

Given:

wavelength 'λ' = 2.4m

pulse width 'τ'= 100T ('T' is the time of one oscillation)

The below inequality express the range of distances to an object that radar can detect

τc/2 < x < Tc/2 ---->eq(1)

Where, τc/2 is the shortest distance

First we'll calculate Frequency 'f' in order to determine time of one oscillation 'T'

f = c/λ (c= speed of light i.e 3 x $10^{8}$ m/s)

f= 3 x $10^{8}$ / 2.4

f=1.25 x $10^{8}$ hz.

As, T= 1/f

time of one oscillation T= 1/1.25 x $10^{8}$

T= 8 x $10^{-9}$ s

It was given that pulse width 'τ'= 100T

τ= 100 x 8 x $10^{-9}$ => 800 x $10^{-9}$ s

From eq(1), we can conclude that the shortest distance to an object that this radar can detect:

$x_{min}$ = τc/2 => (800 x $10^{-9}$ x 3 x $10^{8}$ )/2

$x_{min}$ =120m

8 0

3 years ago