DOMICILE (Choose the word that names a necessary part of the above word.)

Any help please, physics is not my strong suit?

frez [133]

Answer:

2000mg = 2g

5L = 5000mL

16cm = 160mm

5 0

3 years ago

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A student throws a rock horizontally off a 5.0 m tall building. The rock's initial speed is 6.0 m/s. How long will it take the r

Archy [21]

Answer:

The time taken by the rock to reach the ground is 0.569 seconds.

Explanation:

Given that,

A student throws a rock horizontally off a 5.0 m tall building, s = 5 m

The initial speed of the rock, u = 6 m/s

We need to find the time taken by the rock to reach the ground. Using second equation of motion to find it. We get :

$s=ut+\dfrac{1}{2}gt^2\\\\5=6t+\dfrac{1}{2}\times 9.8t^2\\\\t=0.569\ seconds$

So, the time taken by the rock to reach the ground is 0.569 seconds. Hence, this is the required solution.

5 0

3 years ago

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the illuminance on a surface is 6 lux and the surface is 4 meters from the light source. What is the intensity of the saurce?

Lelechka [254]

L = illuminance
A = surface
i = intensity

L = i / A ==: i = L * A

i = 6 lux * 4 m^2 = 24 lumen

8 0

3 years ago

Suppose a car manufacturer tested its cars for front-en4 collisions by hauling them up on a crane and dropping then; from a cert

Brrunno [24]

Answer:

a

Generally from third equation of motion we have that

$v^2 = u^2 + 2a[s_i - s_f]$

Here v is the final speed of the car

u is the initial speed of the car which is zero

$s_i$ is the initial position of the car which is certain height H

$s_i$ is the final position of the car which is zero meters (i.e the ground)

a is the acceleration due to gravity which is g

So

$v^2 = 0 + 2g[H - 0]$

=> $v = \sqrt{ 2 g H}$

b

$H = 9.86 \ m$

Explanation:

Generally from third equation of motion we have that

$v^2 = u^2 + 2a[s_i - s_f]$

Here v is the final speed of the car

u is the initial speed of the car which is zero

$s_i$ is the initial position of the car which is certain height H

$s_i$ is the final position of the car which is zero meters (i.e the ground)

a is the acceleration due to gravity which is g

So

$v^2 = 0 + 2g[H - 0]$

=> $v = \sqrt{ 2 g H}$

When $v = 50 \ km/h = \frac{50 *1000}{3600} = 13.9 \ m/s$ we have that

$13.9 = \sqrt{ 2 g H}$

=> $H = \frac{13.9^2}{2 * 9.8}$

=> $H = 9.86 \ m$

6 0

3 years ago

The density of a hippo is approximately 1030kg/m^3,so it sinks to the bottom of the freshwater lakes and rivers. A 1500kg hippo

hram777 [196]

Answer:

14,700 N

Explanation:

The hyppo is standing completely submerged on the bottom of the lake. Since it is still, it means that the net force acting on it is zero: so, the weight of the hyppo (W), pushing downward, is balanced by the upward normal force, N:

$W-N=0$ (1)

the weight of the hyppo is

$W=mg=(1500 kg)(9.8 m/s^2)=14,700 N$

where m is the hyppo's mass and g is the gravitational acceleration; therefore, solving eq.(1) for N, we find

$N=W=14,700 N$

8 0

3 years ago