a force of 5 newtons accelerates an object the object has a mass of 50 grams what is the objects acceleration

Question 1

galina1969 [7]

1/f= 1/di+1/do
1/f=1/90m+1/40m
1/f ≈27.7m

7 0

3 years ago

A school bus transporting students to school is an example of

jok3333 [9.3K]

Answer:

The answer is D. If I got this right will you please mark me as Brainliest.

3 0

3 years ago

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A horizontal spring is attached to the wall on one end and to a mass on the other end. The mass can slide freely on a frictionle

Radda [10]

Answer:

a) x=0 %T=0, b) x= A %T=100%, c) x=-A %T=50%

Explanation:

This is a simple harmonic movement exercise, which is explained by the expression

x = A cos (wt + Ф)

where angular velocity is related to frequency and period

w = 2π f = 2π / T

we can write the equation of the oscillation

x = A cos θ

When seeing the two equations they are equivalent, so what happens with the angle will also happen with time

We are asked for the percentage of the period at three points: at the maximum elongation and at the point of x = 0, in general the distance is measured from the point of the spring without stretching

The period is defined as the time that the system takes to give a complete oscillation, that is, from x = 0 to x = A and return

a) for the unstretched spring point x = 0

In general, both distance and time are measured from this point, so the percentage of time is zero.

% T = 0

b) for x = A

let's find the angle

cos tea = x / A = 1

therefore the angles tea = 2π rad

when the movement reaches the point of 2π radians it begins to repeat so the period is complete

% T = 100%

c) the point of maximum compression x = -A

let's look for the angles

cos tea = x / A = -1

therefore the angles tea = π rad

at this point the movement is halfway so it should take half the time

% T = 50%

6 0

3 years ago

When a 5.0-kilogram cart moving with a speed of 2.8 meters per second on a horizontal surface collides with a 2.0 kilogram cart

mixer [17]

Here in all such collision type question we can use momentum conservation as we can see that there is no external force on this system

$m_1v_{1i} + m_2v_{2i} = m_1v_{1f} + m_2v_{2f}$

as we know that

$m_1 = 5 kg$

$m_2 = 2 kg$

$v_{1i} = 2.8 m/s$

$v_{2i} = 0 m/s$

now from above equation we have

$5(2.8) + 2(0) = m_1v+ m_2v$

$14 = (5+ 2) v$

$v = 2 m/s$

so the speed of combined system is 2 m/s

8 0

3 years ago

Hank and Harry are two ice skaters whiling away time by playing 'tug of war' between practice sessions. They hold on to opposite

Alex73 [517]

Answer:

the ratio of Hank's mass to Harry's mass is 0.7937 or [ 0.7937 : 1

Explanation:

Given the data in the question;

Hank and Harry are two ice skaters, since both are on top of ice, we assume that friction is negligible.

We know that from Newton's Second Law;

Force = mass × Acceleration

F = ma

Since they hold on to opposite ends of the same rope. They have the same magnitude of force |F|, which is the same as the tension in the rope.

Now,

Mass $_{Hank$ × Acceleration $_{Hank$ = Mass $_{Henry$ × Acceleration $_{Henry$

so

Mass $_{Hank$ / Mass $_{Henry$ = Acceleration $_{Henry$ / Acceleration $_{Hank$

given that; magnitude of Hank's acceleration is 1.26 times greater than the magnitude of Harry's acceleration,

Mass $_{Hank$ / Mass $_{Henry$ = 1 / 1.26

Mass $_{Hank$ / Mass $_{Henry$ = 0.7937 or [ 0.7937 : 1 ]

Therefore, the ratio of Hank's mass to Harry's mass is 0.7937 or [ 0.7937 : 1 ]

8 0

3 years ago