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3 years ago

An object initially at rest experiences an acceleration

Physics

1 answer:

lidiya [134]3 years ago

5 0

Answer:

1.4m/s

Explanation:

Average velocity is the total distance covered divided by the total time taken.

Average velocity = $\frac{total distance }{time }$

Total time taken = 5s + 6s = 11s

The first distance covered = velocity x time = 1.4 x 5 = 7m

second distance covered = velocity x time = 1.4 x 6 = 8.4m

So;

Average velocity = $\frac{7 + 8.4}{11}$ = 1.4m/s

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1. 1500j of work was done to move a box 20m. What force was applied to the box ?

Fantom [35]

Answer:

1. 75N

2. 67,983 J (=67.98 kJ)

Explanation:

1. Work = Force x Distance

we are given that Work = 1,500J and Distance = 20m

hence,

Work = Force x Distance

1,500 = Force x 20

Force = 1,500 ÷ 20 = 75N

2. Potential Energy, PE = mass x gravity x change in height

we are given that mass = 165 kg and change in height = 42m

assuming that gravity, g = 9.81 m/s²

Potential Energy, PE = mass x gravity x change in height

Potential Energy, PE = 165 x 9.81 x 42 = 67,983 J (=67.98 kJ)

4 0

4 years ago

What is the volume of a cone with a height of 27 cm

JulijaS [17]

Explanation:

→ Volume of cone = πr² × h/3

Here,

Radius (r) = 13 cm
Height (h) = 27 cm

→ Volume of cone = π(13)² × 27/3 cm³

→ Volume of cone = 169π × 9 cm³

→ Volume of cone = 1521π cm³

→ Volume of cone = 1521 × 22/7 cm³

→ Volume of cone = 33462/7 cm³

→ <u>Volume of cone = 4780.28 cm³</u>

4 0

3 years ago

In a thunderstorm, electric charge builds up on the water droplets or ice crystals in a cloud. Thus, the charge can be considere

muminat

Answer:

$2.1\cdot 10^{21}$ electrons

Explanation:

The magnitude of the electric field outside an electrically charged sphere is given by the equation

$E=\frac{kQ}{r^2}$

where

k is the Coulomb's constant

Q is the charge stored on the sphere

r is the distance (from the centre of the sphere) at which the field is calculated

In this problem, the cloud is assumed to be a charged sphere, so we have:

$E_b=3.00\cdot 10^6 N/C$ is the maximum electric field strength tolerated by the air before breakdown occurs

$r=1.00 km = 1000 m$ is the radius of the sphere

Re-arranging the equation for Q, we find the maximum charge that can be stored on the cloud:

$Q=\frac{Er^2}{k}=\frac{(3.00\cdot 10^6)(1000)^2}{9\cdot 10^9}=333.3 C$

Assuming that the cloud is negatively charged, then

$Q=-333.3 C$

And since the charge of one electron is

$e=-1.6\cdot 10^{-19}C$

The number of excess electrons on the cloud is

$N=\frac{Q}{e}=\frac{-333.3}{-1.6\cdot 10^{-19}}=2.1\cdot 10^{21}$

5 0

3 years ago

What does the letter I represent?

irina [24]

The letter i is used to signify that a number is an imaginary number. It stand for the square root of negative one.

6 0

2 years ago

Proyektil dengan massa 10 gram ditembak dari senapan dengan massa 5 kg. Kecepatan proyektil sesaat setelah memotret dari laras a

notka56 [123]

Answer:

-0.4 m/s

Explanation:

According to the law of conservation of momentum, the total momentum of the bullet - rifle system must be conserved.

The total momentum before the shot is zero, since they are both at rest:

$p_i = 0$

While the total momentum after the shot can be written as:

$p_f = mv+MV$

where

m = 10 g = 0.010 kg is the mass of the bullet

M = 5 kg is the mass of the rifle

v = 200 m/s is the velocity of the bullet

V is the recoil velocity of the rifle

Since the total momentum is conserved, we can write:

$p_i = p_f$

$0=mv+MV$

And solving for V, we find the recoil velocity:

$V=-\frac{mv}{M}=-\frac{(0.010)(200)}{5}=-0.4 m/s$

and the negative sign indicates that the velocity is opposite to the bullet.

7 0

3 years ago