All categories

2 years ago

PLS ANSWER FAST WILL GIVE BRANLEST!!!

Physics

2 answers:

nevsk [136]2 years ago

7 0

$\huge\bf{\pink{\underline{\underline{\mathcal{AnSwer࿐}}}}}$

<h2>Given:- </h2>

acceleration = 1000 m/s²
force = 5000 N

Mass of the cannonball

<h2>Formula to be used:- </h2>

$\longrightarrow$ $\underline{\boxed{\sf mass = \dfrac{force}{acceleration}}}$

<h2>Solution:- </h2>

: $\implies$ $\sf mass = \dfrac{5000}{1000}$

: $\implies$ $\sf mass = 5 kg$

hence, the required answer is 5kg.

melamori03 [73]2 years ago

4 0

Answer:

mass=force/acceleratio

m=5000/1000

m=5kg

You might be interested in

A plastic rod 1.3 m long is rubbed all over with wool, and acquires a charge of -3e-08 coulombs. we choose the center of the rod

Anestetic [448]

(a) The plastic rod has a length of L=1.3m. If we divide by 8, we get the length of each piece:
$L/8=1.3m/8=0.1625 m$

(b) The center of the rod is located at x=0. This means we have 4 pieces of the rod on the negative side of x-axis, and 4 pieces on the positive side. So, starting from x=0 and going towards positive direction, we have: piece 5, piece 6, piece 7 and piece 8. Each piece is 0.1625 m long. Therefore, the center of piece 5 is at 0.1625m/2=0.0812 m. And the center of piece 6 will be shifted by 0.1625m with respect to this:
$c_6 = 0.0812m+0.1625m=0.2437 m$

(c) The total charge is $Q=-3 \cdot 10^{-8}C$ . To get the charge on each piece, we should divide this value by 8, the number of pieces:
$Q/8=-3\cdot 10^{-8}C/8=-3.75\cdot 10^{-9}C$

(d) We have to calculate the electric field at x=0.7 generated by piece 6. The charge on piece 6 is the value calculated at point (c):
$q= -3.75\cdot 10^{-9}C$
If we approximate piece 6 as a single charge, the electric field is given by
$E=k_e \frac{q}{d^2}$
where $k_e=8.99\cdot 10^9Nm^2C^{-2}$ and d is the distance between the charge (center of piece 6, located at 0.2437m) and point a (located at x=0.7m). Therefore we have
$E= 8.99\cdot 10^9 Nm^2C^{-2} \frac{-3.75\cdot 10^9 C}{(0.2437m-0.7m)^2} =-161.9 V/m$
poiting towards the center of piece 6, since the charge is negative.

(e) missing details on this question.

5 0

2 years ago

Does exists the friction force in space?If yes,tell an full example

VashaNatasha [74]

Yes, friction does exist in space. Friction has nothing to do with the earth's atmosphere. It exists everywhere in the universe. <span />

5 0

3 years ago

If you are describing the displacement of an object with respect to a reference point, you should use

Leya [2.2K]

You should use B. a vector

6 0

2 years ago

People hoping to travel to other worlds are faced with huge challenges. One of the biggest is the time required for a journey. T

Elena L [17]

Answer:

It would take 8.22037 hrs away. Wouldn't it?

Explanation:

Because

4.11016

= 8.22037

5 0

3 years ago

After falling from rest at a height of 28.7 m, a 0.502 kg ball rebounds upward, reaching a height of 19.8 m. If the contact betw

Ad libitum [116K]

Answer:

9080 N

Explanation:

Consider the two motions of the ball.

In the downward motion, initial velocity, <em>u</em>, is 0 (because it falls from rest) and the distance is 28.7 m. Using the equation of motion and using <em>g</em> as 9.8 m/s²,

<em>v² = </em>0² + 2 × 9.8 × 28.7<em> </em>= 562.52

<em>v = </em>19.7 m/s

<em />

For the downward motion, the initial velocity is unknown, the final velocity is 0 and initial velocity is desired. <em>g</em> is negative because the motion is upwars.

<em>0² = v² - </em>2 × 9.8 × 19.8

<em>v² = </em>388.08

<em>v = </em>10.7 m/s

The change in momentum = 0.502(10.7 -(23.7)) = 21.7868 kgm/s

The impulse = change in monetum

Ft = 21.7868 kgm/s

But t = 2.4 ms

$[tex]F = \dfrac{21.7868}{2.4\times10{-3}} = 9078 \text{ N}$ [\tex]

8 0

3 years ago