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3 years ago

Hitungkan pecutan bagi blok di bawah: / Cal(a)m= 2 kgF= 8.0 N

Physics

1 answer:

ioda3 years ago

4 0

Answer:

Acceleration = 4 m/s²

Explanation:

Given the following data;

Force = 8 N

Mass = 2 kg

To find the acceleration of the block;

Newton's Second Law of Motion states that the acceleration of a physical object is directly proportional to the net force acting on the physical object and inversely proportional to its mass.

Mathematically, it is given by the formula;

$Acceleration = \frac {Net \; force}{mass}$

Substituting into the formula, we have;

$Acceleration = \frac {8}{2}$

Acceleration = 4 m/s²

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A +2e charge is at the point (-1,0) mm in the x,y plane. A –e charge is at the point (0,1) mm. What is the electric field at the

Gennadij [26K]

Answer:

Let I and j be the unit vector along x and y axis respectively.

Electric field at origin is given by

E= kq1/r1^2 i + kq2/r2^2j

= 9*10^9*1.6*10^-19*/10^-6*(2i+ j)

= (2.88i + 1.44j)*10^-3 N/C

Force on charge= qE= 3*10^-19*1.6*(2.88i +1. 44 j) *10^-3

F= (1.382 i + 0.691 j) *10^-21

Goodluck

Explanation:

4 0

2 years ago

A commuter backs her car out of her garage with an acceleration of 1.40 m/s2.A) How long does it take her to reach a speed of 2.

AURORKA [14]

Answer:

(A) 1.43secs

(B) -2.50m/s^2

Explanation:

A commuter backs her car out of her garage with an acceleration of 1.40m/s^2

(A) When the speed is 2.00m/s then, the time can be calculated as follows

t= Vf-Vo/a

The values given are a= 1.40m/s^2 , Vf= 2.00m/s, Vo= 0

= 2.00-0/1.40

= 2.00/1.40

= 1.43secs

(B) The deceleration when the time is 0.800secs can be calculated as follows

a= Vf-Vo/t

= 0-2.00/0.800

= -2.00/0.800

= -2.50m/s^2

4 0

3 years ago

Alex is asked to move two boxes of books in contact with each other and resting on a rough floor. He decides to move them at the

Over [174]

Answer:

Part a)

$a= 0.32 m/s^2$

Part b)

$F_c = 3.6 N$

Part c)

$F_c = 5.5 N$

Explanation:

Part a)

As we know that the friction force on two boxes is given as

$F_f = \mu m_a g + \mu m_b g$

$F_f = 0.02(10.6 + 7)9.81$

$F_f = 3.45 N$

Now we know by Newton's II law

$F_{net} = ma$

so we have

$F_p - F_f = (m_a + m_b) a$

$9.1 - 3.45 = (10.6 + 7) a$

$a = \frac{5.65}{17.6}$

$a= 0.32 m/s^2$

Part b)

For block B we know that net force on it will push it forward with same acceleration so we have

$F_c - F_f = m_b a$

$F_c = \mu m_b g + m_b a$

$F_c = 0.02(7)(9.8) + 7(0.32)$

$F_c = 3.6 N$

Part c)

If Alex push from other side then also the acceleration will be same

So for box B we can say that Net force is given as

$F_p - F_f - F_c = m_b a$

$9.1 - 0.02(7)(9.8) - F_c = 7(0.32)$

$F_c = 9.1 - 0.02 (7)(9.8) - 7(0.32)$

$F_c = 5.5 N$

3 0

3 years ago

when a hockey stick strike a hockey puck and sends it towards the goalie, (thermal, mechanical) energy is being transferred from

velikii [3]

Answer:

Yes... definitely true.. can you tell me what is the question..

Explanation:

Please mark me brainliest :D

And also.. I would appreciate it if you give me 5 stars and a thanks.. :D

8 0

2 years ago

Select all that apply.

ikadub [295]

All are true except the first two.

6 0

3 years ago

Read 2 more answers

Hitungkan pecutan bagi blok di bawah: / Cal(a)m= 2 kgF= 8.0 N​

Hitungkan pecutan bagi blok di bawah: / Cal(a)m= 2 kgF= 8.0 N