Need help can anyone solve this for me ?

A 4.5-kg object oscillates on a horizontal spring with an amplitude of 3.8 cm. Its maximum acceleration is 26 m/s2 . Find (a) th

zimovet [89]

Answer:

F = - K X force constant for spring

a = F / m maximum acceleration

F = 4.5 kg * 26 m/s^2 = 117 Newtons

(A) K = 117 N / .038 m = 3079 N/m

ω = (K/M)^1/2 = (117/5)^1/2 = 4.84 / sec

(B) f = ω / 2 pi = 4.84 / 6.28 = .77 /sec

(C) P = 1 / f = 1/.77 = 1.30 sec

3 0

2 years ago

A weight lifter picks up a barbell and 1. lifts it chest high 2. holds it for 30 seconds 3. puts it down slowly (but does not dr

Anna11 [10]

1. lifts it chest high

The force opposing to this action is the force due to gravity. Therefore the work done is:

W1 = m g d

where m is mass of the barbell, g is gravity and d is displacement

2. holds it for 30 seconds

Work is a product of force and displacement, since there is no displacement, therefore work done is zero.

W2 = 0

3. puts it down slowly

If the barbell was dropped, then it would simply be a free fall. But since it was not, so the work done here is also equal to the weight of the barbell times displacement:

W3 = m g d

We can see that W1 = W3, and since W2 = 0, therefore the answer is:

7 0

2 years ago

The shape of the earth is an _______ ________

Nastasia [14]

The shape of the earth is an <span>oblate spheroid.</span>

3 0

3 years ago

Force F acts between a pair of charges, q1 and q2, separated by a distance d. For each of the statements, use the drop-down menu

lora16 [44]

The initial force between the two charges is given by:

$F=k \frac{q_1 q_2}{d^2}$

where k is the Coulomb's constant, q1 and q2 the two charges, d their separation. Let's analyze now the other situations:

1. F

In this case, q1 is halved, q2 is doubled, but the distance between the charges remains d.

So, we have:

$q_1' = \frac{q_1}{2}\\q_2' = 2 q_2\\d' = d$

So, the new force is:

$F'=k \frac{q_1' q_2'}{d'^2}= k \frac{(\frac{q_1}{2})(2q_2)}{d^2}=k \frac{q_1 q_2}{d^2}=F$

So the force has not changed.

2. F/4

In this case, q1 and q2 are unchanged. The distance between the charges is doubled to 2d.

So, we have:

$q_1' = q_1\\q_2' = q_2\\d' = 2d$

So, the new force is:

$F'=k \frac{q_1' q_2'}{d'^2}= k \frac{q_1 q_2)}{(2d)^2}=\frac{1}{4} k \frac{q_1 q_2}{d^2}=\frac{F}{4}$

So the force has decreased by a factor 4.

3. 6F

In this case, q1 is doubled and q2 is tripled. The distance between the charges remains d.

So, we have:

$q_1' = 2 q_1\\q_2' = 3 q_2\\d' = d$

So, the new force is:

$F'=k \frac{q_1' q_2'}{d'^2}= k \frac{(2 q_1)(3 q_2)}{d^2}=6 k \frac{q_1 q_2}{d^2}=6F$

So the force has increased by a factor 6.

8 0

2 years ago

Read 2 more answers

What forces contribute to density

mixer [17]

No force contributes to density, the density is a physical quantity that is defined as being $\rho = m/V 
$ the raport between the mass of the object and its volume. However if you want to measure the density of an object you might want to determine its gravity force (weight) $G=mg$ from which knowing the gravitational acceleratin you can find its mass $m=G/g
$
where $G$ is given in Newtons and $g=9.81$ is given in $(m/s^2)
$

5 0

2 years ago