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3 years ago

HELP ASAP IM GIVING BRAINLIEST PLSSS

Physics

2 answers:

klemol [59]3 years ago

6 0

Answer:

The first picture's area of least repulsion is the areas of the outer portions where the souths are pointing. The second picture's area of least repulsion is the area in the inner part

Explanation:

IgorC [24]3 years ago

5 0

Answer:

In the first image, the areas with the least repulsion force are the blue squares next to the boxes with the letter "S". In the second image, the area with the least repulsion force is the square between the boxes with the letter "S" and "N".

Explanation:

As we can see, the images above represent magnets. Magnets work on the basis of magnetism, which is the phenomenon of attraction or repulsion between the poles of these magnets. Approaching the north pole of one magnet and the south pole of another magnet, an attraction is noted. However, approaching two equal poles as if repulsion is attempted. In short, poles of the same nature repel each other and of different natures are attracted.

With that, we can conclude, through the image above that in the first image, the areas with the least repulsion force are the blue squares next to the boxes with the letter "S". In the second image, the area with the least repulsion force is the square between the boxes with the letter "S" and "N".

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An 80.0 kg hiker walks a distance of 400.0 m along a road that slopes 5.0 degrees upward, and then stops. What is the hiker's fi

lana66690 [7]

The height difference is found by
$\delta H=400sin(5 \°)=34.86m$
Then the change in potential energy is
$E=mgh=(80.0kg)(9.8 \frac{kgm}{s^2})(34.86)= 27332J$

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3 years ago

A spring with spring constant 15.5 N/m hangs from the ceiling. A ball is suspended from the spring and allowed to come to rest.

Aloiza [94]

Answer:

A) 138.8g

B)73.97 cm/s

Explanation:

K = 15.5 Kn/m

A = 7 cm

N = 37 oscillations

tn = 20 seconds

A) In harmonic motion, we know that;

ω² = k/m and m = k/ω²

Also, angular frequency (ω) = 2π/T

Now, T is the time it takes to complete one oscillation.

So from the question, we can calculate T as;

T = 22/37.

Thus ;

ω = 2π/(22/37) = 10.5672

So,mass of ball (m) = k/ω² = 15.5/10.5672² = 0.1388kg or 138.8g

B) In simple harmonic motion, velocity is given as;

v(t) = vmax Sin (ωt + Φ)

It is from the derivative of;

v(t) = -Aω Sin (ωt + Φ)

So comparing the two equations of v(t), we can see that ;

vmax = Aω

Vmax = 7 x 10.5672 = 73.97 cm/s

6 0

3 years ago

On January 22, 1943, the temperature in Spearfish, South Dakota, rose from -4.0∘F∘F to 45.0∘F∘F in just 2 minutes. What was the

Margaret [11]

Answer:

The change in temperature, $\Delta T=9.45^{\circ} C$

Explanation:

Given that,

The temperature in Spearfish, South Dakota, rose from $-4^{\circ} F\ to\ 45^{\circ} F$ in just 2 minutes. We need to find the temperature change in Celsius degrees. Change in temperature is given by final temperature minus initial temperature such that,

$\Delta T=T_f-T_i\\\\\Delta T=45-(-4)\\\\\Delta T=49^{\circ}F$

The relation between degrees Celsius and degrees Fahrenheit is given by :

$F=1.8C+32$

Here, F = 49 degrees

$49=1.8C+32\\\\\Delta T=9.45^{\circ} C$

So, the change in temperature is 9.45 degree Celsius. Hence, this is the required solution.

8 0

3 years ago

Problem A spring was at its resting position where it is attached to a wall at its left side and a block at its right side as sh

puteri [66]

Answer:

F = 19.1 N

Explanation:

To find the force exerted by the string on the block you use the following formula:

$F=kx$ (1)

k: spring constant = 95.5 N/m

x: displacement of the block from its equilibrium position = 0.200 m

you replace the values of k and x in the equation (1):

$F=(95.5N/m)(0.200m)=19.1N$

Hence, the force exterted on the block is 19.1 N

8 0

3 years ago

A cart moving at 10 m/s is brought to a stop by the force plotted in the force-time graph shown here. Find the impulse and the a

Elena L [17]

Answer:

Impulse = 88 kg m/s

Mass = 8.8 kg

Explanation:

<u>We are given a graph of Force vs. Time. Looking at the graph we can see that the Force acts approximately between the time interval from 1sec to 4sec. </u>

Newton's Second Law relates an object's acceleration as a function of both the object's mass and the applied net force on the object. It is expressed as:

$F=ma$ Eqn. (1)

where

$F$ : is the Net Force in Newtons ( $N$ )

$m$ : is the mass ( $kg$ )

$a$ : is the acceleration ( $m/s^2$ )

We also know that the acceleration is denoted by the velocity ( $v$ ) of an object as a function of time ( $t$ ) with

$a=\frac{v}{t}$ Eqn. (2)

Now substituting Eqn. (2) into Eqn. (1) we have

$F=m\frac{v}{t}\\ \\Ft=mv$ Eqn. (3)

However since in Eqn. (3) the time-variable is present, as a result the left hand side (i.e. $Ft$ is in fact the Impulse $J$ of the cart ), whilst the right hand side denotes the change in momentum of the cart, which by definition gives as the impulse. Also from the graph we can say that the Net Force is approximately ≈ $22N$ and $t=4 sec$ (thus just before the cut-off time of the force acting).