Does anyone know 08fortnitebeast?

The figure above shows the net force exerted on an object as a function of the position of the object. The object starts from re

weqwewe [10]

Answer:

0.06 Kg

Explanation:

From the question given above, the following data were obtained:

Initial velocity (u) = 0 m/s

Final velocity (v) = 3.0 m/s

Distance (s) = 0.09 m

Net Force (F) = 3 N

Mass (m) =?

Next, we shall determine the acceleration of the object. This can be obtained as follow:

Initial velocity (u) = 0 m/s

Final velocity (v) = 3.0 m/s

Distance (s) = 0.09 m

Acceleration (a) =?

v² = u² + 2as

3² = 0² + (2 × a × 0.09)

9 = 0 + 0.18a

9 = 0.18a

Divide both side by 0.18

a = 9 / 0.18

a = 50 m/s²

Finally, we shall determine the mass of the object. This can be obtained as follow:

Net Force (F) = 3 N

Acceleration (a) = 50 N

Mass (m) =?

F = ma

3 = m × 50

Divide both side by 50

m = 3 / 50

m = 0.06 Kg

Therefore, the mass of the object is 0.06 Kg

5 0

2 years ago

I'd like you to think back on 2 events in

strojnjashka [21]

Answer:

Holi

biwali

these are best events and love to celebrate withmy family and friends it contains lot of happiness and joy

7 0

2 years ago

A child whirls a 3.00 kg ball on a string .50 m from the axis of rotation in a horizontal circle. The ball makes 1 revolution in

melamori03 [73]

Answers:

a) $0.5 m/s^{2}$

b) $1.5 N$

Explanation:

a) The centripetal acceleration $a_{c}$ of an object moving in a uniform circular motion is given by the following equation:

$a_{c}=\omega^{2} r$

Where:

$\omega=1 \frac{rev}{s}$ is the angular velocity of the ball

$r=0.5 m$ is the radius of the circular motion, which is equal to the length of the string

Then:

$a_{c}=(1 \frac{rev}{s})^{2} 0.5 m$

$a_{c}=0.5 m/s^{2}$ This is the centripetal acceleration of the ball

b) On the other hand, in this circular motion there is a force (centripetal force $F$ ) that is directed towards the center and is equal to the tension ( $T$ ) in the string:

$F=T=m. a_{c}$

Where $m=3 kg$ is the mass of the ball

Hence:

$T=(3 kg)(0.5 m/s^{2})$

$T=1.5 N$ This is the tension in the string

7 0

3 years ago

A force of 6.00 N acts in the positive direction on a 3.00 kg object, originally traveling at +15.0 m/s, for 10.0 s. (a) What is

frozen [14]

Answer:

60 kg m/s

Explanation:

Let $a\;\; m/s^2$ be the acceleration of the object.

As the acceleration of the object is constant, so

$a=\frac {v-u}{t}\cdots(i)$

Given that applied force, F=6.00 N,

From Newton's second law, we have

$F= m\times a$ ,

$\Rightarrow F=\frac {m(v-u)}{t}$ [from equation (i)]

$\Rightarrow Ft=m(v-u)$

$\Rightarrow Ft=mv-mu$

$\Rightarrow mv-mu=6\times 10$ [given that time, t=10 s and F=6 N]

$\Rightarrow mv-mu=60 kg \;m/s$

Here mv is the final momentum of the object and mu is the initial momentum of the object.

So, the change in the momentum of the object is mv-mu.

Hence, the change in the momentum of the object is 60 kg m/s.

6 0

3 years ago

a ball is thrown downward with an initial speed of 7 m/s. the ball's velocity after 3 seconds is ____ m/s (g= -9.8m/s^{2}

Karo-lina-s [1.5K]

The speed downwards is 7 + (3*9.8)
7+ 29.4 = 36.4 m/s

5 0

2 years ago