All categories

3 years ago

A rock exerts 5000 Pa of pressure on the ground. If the rock weighs 250 N, how much area is in contact with the ground?

Physics

2 answers:

ololo11 [35]3 years ago

4 0

P = F/S - S = F/P = 250/5000= 0.05 m2

SOVA2 [1]3 years ago

3 0

Answer:

Area=0.05 $m^{2}$

Explanation:

Given pressure=5000 Pa

and weight=250 N

Using equation

Pressure=Force/Area

So,Area=Force/pressure

Area=250/5000

Area=0.05 $m^{2}$

You might be interested in

Can you be fluid overload and dehydrated at the same time

STALIN [3.7K]

.................yes...................

5 0

3 years ago

Which description correctly summarizes how an electric motor causes an axle to turn?

alexira [117]

Answer:

B on edge

Explanation:

the piles of the magnetic field generated around the armature are attracted to the opposite poles of the permanent magnent. As the opposite poles align, the commutator reverses the current direction so like poles are aligned and the armature continues to spin

6 0

3 years ago

Please help me:(

Lelechka [254]

Answer:B

Explanation:

I’m doing the same thing you are!

3 0

3 years ago

You're driving down the highway late one night at 20 m/s when a deer steps onto the road 44 m in front of you. Your reaction tim

never [62]

Answer:

14.0 m

25.1 m/s

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

Distance traveled in the reaction time

Distance = Speed × Time

$\text{Distance}=20\times 0.5=10\ m$

$v^2-u^2=2as\\\Rightarrow s=\frac{v^2-u^2}{2a}\\\Rightarrow s=\frac{0^2-20^2}{2\times -10}\\\Rightarrow s=20\ m$

Distance in which the car will stop is 10+20 = 30.0 m

So, the car will not hit the deer

Distance between the car and deer is 44-30 = 14.0 m

$\text{Distance}=u\times 0.5=0.5u\ m$

$v^2-u^2=2as\\\Rightarrow -u^2=2as-v^2\\\Rightarrow u^2=v^2-2as\\\Rightarrow u^2=0^2-2\times -10\times (44-0.5u)\\\Rightarrow u^2=20(44-0.5u)\\\Rightarrow u^2=880-10u\\\Rightarrow u^2+10u-880=0$

$u=\frac{-10+\sqrt{10^2-4\cdot \:1\left(-880\right)}}{2\cdot \:1}, u=\frac{-10-\sqrt{10^2-4\cdot \:1\left(-880\right)}}{2\cdot \:1}\\\Rightarrow u=25.08, -35.08\ m/s$