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kondaur [170]
3 years ago
15

Brianna pushes a 20 kg box with a force of 50N to achieve an acceleration of 2.5 m/s/s. In order to push a 30 kg box at the same

acceleration, how much force will Brianna need to apply?
Physics
1 answer:
Elanso [62]3 years ago
5 0

Answer:

She applies 75N of force

Explanation:

F=ma

50=20(2.5)

F=30(2.5)

F=75 N

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A ball is spun around in circular motion such that it completes 50 rotations in 25 s.
Leokris [45]

Answer:

(A) The period of its rotation is 0.5 s (2) The frequency of its rotation is 2 Hz.

Explanation:

Given that,

a ball is spun around in circular motion such that it completes 50 rotations in 25 s.

(1). Let T be the period of its rotation. It can be calculated as follows :

T=\dfrac{25}{50}\\\\T=0.5\ s

(2). Let f be the frequency of its rotation. It can be defined as the number of rotations per unit time. So,

f=\dfrac{50}{25}\\\\f=2\ Hz

Hence, this is the required solution.

3 0
3 years ago
A water wheel rotates with a period of 2.26 s. If the water wheel has a radius of l.94 m,
Dafna11 [192]

Answer:

The correct answer is B)

Explanation:

When a wheel rotates without sliding, the straight-line distance covered by the wheel's center-of-mass is exactly equal to the rotational distance covered by a point on the edge of the wheel.  So given that the distances and times are same, the translational speed of the center of the wheel amounts to or becomes the same as the rotational speed of a point on the edge of the wheel.

The formula for calculating the velocity of a point on the edge of the wheel is given as

V_{r} = 2π r / T

Where

π is Pi which mathematically is approximately 3.14159

T is period of time

Vr is Velocity of the point on the edge of the wheel

The answer is left in Meters/Seconds so we will work with our information as is given in the question.

Vr = (2 x 3.14159 x 1.94m)/2.26

Vr = 12.1893692/2.26

Vr = 5.39352619469

Which is approximately 5.39

Cheers!

7 0
3 years ago
A 6.8 kg bowling ball and 7.4 kg bowling ball rest on a rack 0.74 m apart. What is the force of gravity pulling each ball toward
zimovet [89]

The gravitational force between the two balls is 6.13\cdot 10^{-9} N

Explanation:

The magnitude of the gravitational force between two objects is given by:

F=G\frac{m_1 m_2}{r^2}

where :

G=6.67\cdot 10^{-11} m^3 kg^{-1}s^{-2} is the gravitational constant

m1, m2 are the masses of the two objects

r is the separation between the objects

For the balls in this problem,  we have

m_1 = 6.8 kg

m_2 = 7.4 kg

r = 0.74 m

Substituting into the equation, we find the gravitational force between the two balls:

F=(6.67\cdot 10^{-11})\frac{(6.8)(7.4)}{(0.74)^2}=6.13\cdot 10^{-9}N

Learn more about gravitational force:

brainly.com/question/1724648

brainly.com/question/12785992

#LearnwithBrainly

4 0
3 years ago
Me ajudem, Por favor!!!!!!
zzz [600]

Answer:

a)  a=4\,\frac{m}{s^2}

b)  V(t)=4\,t\,+3

c)  V(1)=7 \,\frac{m}{s} \\

d)  Displacement = 22 m

e)  Average speed = 11 m/s

Explanation:

a)

Notice that the acceleration is the derivative of the velocity function, which in this case, being a straight line is constant everywhere, and which can be calculated as:

slope= \frac{15=3}{3-0} =4\,\frac{m}{s^2}

Therefore,  acceleration is a=4\,\frac{m}{s^2}

b) the functional expression for this line of slope 4 that passes through a y-intercept at (0, 3) is given by:

y=m\,x+b\\V(t)=4\,t\,+3

c) Since we know the general formula for the velocity, now we can estimate it at any value for 't", for example for the requested t = 1 second:

V(t)= 4\,t+3\\V(1)=4\,(1)+3\\V(1)=7 \,\frac{m}{s}

d) The displacement between times t = 1 sec, and t = 3 seconds is given by the area under the velocity curve between these two time values. Since we have a simple trapezoid, we can calculate it directly using geometry and evaluating V(3) (we already know V(1)):

Displacement = \frac{(7+15)\,2}{2} = 22\,\,m

e) Recall that the average of a function between two values is the integral (area under the curve) divided by the length of the interval:

Average velocity = \frac{22}{2} = 11\, \,\frac{m}{s}

3 0
3 years ago
How far does a car travel in 90 seconds if it’s traveling 55 m/s? Show equation
podryga [215]

You just said the car is traveling at the speed of 55 m/s.  If I understand this  correctly, that means the car will cover:

55 meters in the first second,

55 meters in the 2nd second,

55 meters in the 3rd second,

55 meters in the 4th second,

55 meters in the 5th second,

.

.

.

55 meters in the 87th second,

55 meters in the 88th second,

55 meters in the 89th second, and

55 meters in the 90th second.

That's 55 meters 90 times.  If you just move these words around a little bit, it says "90 times 55 meters" . . . a pretty simple arithmetic problem.

The equation is . . . <em>Distance = (55 m/s) times (time, in seconds)</em> .

I get <em>4,953 meters</em>.  You should check me on this.

8 0
3 years ago
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