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Vanyuwa [196]
4 years ago
5

Earth attracts a person with a gravitational force of 7.0 × 102 newtons. What is the magnitude of the force with which the indiv

idual attracts Earth? ..
A)3.0 × 102 newtons
B)7.0 × 102 newtons
C)8.5 × 102 newtons
D)9.8 × 102 newtons
Physics
2 answers:
Agata [3.3K]4 years ago
6 0

Answer:

B.  7.0 × 102 newtons

Explanation:

kirza4 [7]4 years ago
3 0
B. In the system of the object and the earth, by Newton's third law, the object will exert and equal and opposite force upon the earth.
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You have a stopped pipe of adjustable length close to a taut 62.0 cmcm, 7.25 gg wire under a tension of 4710 NN. You want to adj
Oksana_A [137]

Answer:

6 cm long

Explanation:

F = 4110N

Vo(speed of sound) = 344m/s

Mass = 7.25g = 0.00725kg

L = 62.0cm = 0.62m

Speed of a wave in string is

V = √(F / μ)

V = speed of the wave

F = force of tension acting on the string

μ = mass per unit density

F(n) = n (v / 2L)

L = string length

μ = mass / length

μ = 0.00725 / 0.62

μ = 0.0116 ≅ 0.0117kg/m

V = √(F / μ)

V = √(4110 / 0.0117)

v = 592.69m/s

Second overtone n = 3 since it's the third harmonic

F(n) = n * (v / 2L)

F₃ = 3 * [592.69 / (2 * 0.62)

F₃ = 1778.07 / 1.24 = 1433.927Hz

The frequency for standing wave in a stopped pipe

f = n (v / 4L)

Since it's the first fundamental, n = 1

1433.93 = 344 / 4L

4L = 344 / 1433.93

4L = 0.2399

L = 0.0599

L = 0.06cm

L = 6cm

The pipe should be 6 cm long

6 0
3 years ago
An ant crawls along a sidewalk with a velocity of 0.1 m/s in a direction that is 45° relative to edge of the sidewalk. If it has
Marat540 [252]

Answer:14 s

Explanation:

Given

Velocity of ant is 0.1 m/s in a direction 45^{\circ}

if it has traveled 1 m perpendicular to the edge of the sidewalk

i.e. from diagram

\sin 45=\frac{1}{L}

L=\sqrt{2}

time=\frac{Distance}{speed}

t=\frac{\sqrt{2}}{0.1}

t=14.14 s

3 0
3 years ago
By counting the number of crests that pass in a given amount of time, a person can calculate the
ladessa [460]
By calculating the crests, you can find the waves' frequency.
Hope this helps!
3 0
3 years ago
A cart starting from rest rolls down a frictionless 1.0 m high hill. It travels a distance 2.0 m along the rough bottom surface
Rzqust [24]

Answer:

Option B is correct.

Explanation:

Given data

Height of the hill = AB = 1 m

Distance traveled  along the rough bottom surface = AC = 2 m

Now from the ΔABC

\sin \theta = \frac{AB}{AC}

\sin \theta = \frac{1}{2}

\theta = 30 °

We know that  the coefficient of kinetic friction is

\mu = \tan \theta

\mu = \tan 30

\mu = 0.5

This is the value of the coefficient of kinetic friction

Thus option B is correct.

8 0
3 years ago
A motorboat traveling from one shore to the other at a rate of 5m/s east encounters a current flowing at a rate of 3.5m/s north
nikdorinn [45]

Answer:

Resultant velocity will be equal to 6.10 m/sec

Explanation:

We have given a motorbike is traveling with 5 m/sec in east

And a current is flowing at a rate of 3.5 m /sec in north

We know that east and north is perpendicular to each other

So resultant velocity will be vector sum of both velocity

So resultant velocity v=\sqrt{5^2+3.5^2}=6.10m/sec

So resultant velocity will be equal to 6.10 m/sec

6 0
4 years ago
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