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3 years ago

The graph above represents an object moving with a __________________.

Physics

2 answers:

densk [106]3 years ago

3 0

... cloak of invisibility spread over it.

Firlakuza [10]3 years ago

3 0

D) constant negative acceleration

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If the hiker starts climbing at an elevation of 350 ft, what will their change in gravitational potential energy be, in joules,

Kitty [74]

Answer:

352,088.37888Joules

Explanation:

Complete question;

A hiker of mass 53 kg is going to climb a mountain with elevation 2,574 ft.

A) If the hiker starts climbing at an elevation of 350 ft., what will their change in gravitational potential energy be, in joules, once they reach the top? (Assume the zero of gravitational potential is at sea level)

Chane in potential energy is expressed as;

ΔGPH = mgΔH

m is the mass of the hiker

g is the acceleration due to gravity;

ΔH is the change in height

Given

m = 53kg

g = 9.8m/s²

ΔH = 2574-350 = 2224ft

since 1ft = 0.3048m

2224ft = (2224*0.3048)m = 677.8752m

Required

Gravitational potential energy

Substitute the values into the formula;

ΔGPH = mgΔH

ΔGPH = 53(9.8)(677.8752)

ΔGPH = 352,088.37888Joules

Hence the gravitational potential energy is 352,088.37888Joules

7 0

3 years ago

Jorge wants to know if yearly income and amount of tax returns are related in any way. In looking for an association between

emmainna [20.7K]

The answer will be b

4 0

3 years ago

Two point charges, 3.4 μC and -2.0 μC , are placed 5.0 cm apart on the x axis. Assume that the negative charge is at the origin,

Elza [17]

The electric field is zero at x = -16.45cm

Data;

q1 = 3.4 μC
q2 = -2.0 μC
distance = 5cm

<h3>The Electric Field at point 0</h3>

As the 3μC is larger than -2.0μC and the charges are opposite sign. The electric field will be zero at the negative axis.

Let the point be at x.

For an electric field to be equal to zero;

$k(\frac{q_1}{d_1})^2 + k(\frac{q_2}{d_2})^2 = 0\\\frac{3.4}{(5-x)^2} - \frac{2}{x^2} = 0\\$

Let's solve for x using mathematical methods.

$\frac{3.4x^2 - 2(5-x)^2}{(5-x)^2x^2}= 0\\ 3.4x^2 - 2(5-x^2) = 0\\3.4x^2 - 50 - 2x^2 + 20x = 0\\1.4x^2 +20x - 50 = 0$

Solving the above quadratic equation;

$x = -16.45cm$

The electric field is zero at x = -16.45cm

Learn more on electric field at a point here;

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8 0

3 years ago

A curent-carrying 2.0-cm long segment of wire is inside a long solenoid adius - 40 cm - 800 turns/m, current - 50 mA). The wire

klasskru [66]

Answer:

The value is $F_{max} = 12 \muN$

Explanation:

From the question we are told that

The length is $l = 2.0 \ cm = 0.02 \ m$

The radius of the the solenoid is $r =40 \ cm$

The number of turns per meter is $N = 800 \ turns / m$

The current through the solenoid is $I = 50 \ mA = 50*10^{-3} \ A$

The current through the segment is $I_s = 12 \ A$

Generally the magnetic force is mathematically represented as

$F = B * I_s l sin(\theta)$

At maximum $\theta = 90^o$

$F_{max} = B * I_s l$

Here B is the magnetic field is mathematically represented as

$B = \mu_o * N * I$

Here $\mu_o$ is the permeability of free space with value

$\mu_o = 4\pi * 10^{-7} N/A^2$

$B = 4\pi * 10^{-7} * 800 *50*10^{-3}$

=> $B = 5.0272 *10^{-5} \ T$

$F_{max} = 5.0272 *10^{-5} * 12 * 0.02$

$F_{max} = 1.2 *10^{-5}$

$F_{max} = 12 \muN$

4 0

3 years ago

How much force will it take to lift a 54n object if it is on a lever that is 19m long on the side the object is on while i push

avanturin [10]

Answer:

How long will it take to travel a distance of 96 km? Givens. Solving For ... An object accelerates 3.0 m/s2 when a force of 6.0 Newtons is applied to it.

Explanation:

6 0

3 years ago