What word is used to describe an acid or base that can destroy body tissue, clothing, and many other?

1 answer:

7 0

Corrosive. It’s something that tends to cause corrosion, and it means to destroy or damage things slowly by chemical action.

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What if m1 is initially moving at 3.4 m/s while m2 is initially at rest? (a) find the maximum spring compression in this case?

Lisa [10]

<span>Ans : Initial E = KE = Â˝mvÂ˛ = Â˝ * 1.2kg * (2.2m/s)Â˛ = 2.9 J max spring compression where both velocities are the same: conserve momentum: 1.2kg * 2.2m/s = (1.2 + 3.2)kg * v â†’ v = 0.6 m/s which means the combined KE = Â˝ * (1.2 + 3.2)kg * (0.6m/s)Â˛ = 0.79 J The remaining energy went into the spring: U = (2.9 - 0.79) J = 2.1 J = Â˝kxÂ˛ = Â˝ * 554N/m * xÂ˛ x = 0.0076 m â† (a)</span>

7 0

3 years ago

The skateboarder has a mass of 100 kg. When traveling downward from E to D, he reaches a velocity of 11 m/s. Calculate his kinet

patriot [66]

6050 J is the kinetic energy at D

<u>Explanation:</u>

In physics, the object's kinetic energy (K.E) defined as the energy it possesses during movement. It can be defined as the required work to accelerate a certain body weight in order to rest at a certain speed. When the body receives this energy as it speeds up (accelerates), it retains this energy unless speed varies. The equation is given as,

$K . E=\frac{1}{2} \times m \times v^{2}$

Where,

m - mass of an object

v - velocity of the object

Here,

Given data:

m = 100 kg

v = 11 m/s

By substituting the given values in the above equation, we get

$K . E=\frac{1}{2} \times 100 \times(11)^{2}=\frac{1}{2} \times 100 \times 121=\frac{12100}{2}=6050\ \mathrm{J}$

6 0

3 years ago

sara and tory are out fishing on the lake on a hot summer day when they both decide to go for a swim. sara dives off the front o

crimeas [40]

Here it is an application of Newton's III law

as we know by Newton's III law that every action has equal and opposite reaction

So here as we know that two boys jumps off the boat with different forces

from front side of the boat the boy jumps off with force 45 N which means as per Newton's III law if boy has a force of 45 N in forward direction then he must apply a reaction force on the boat in reverse direction of same magnitude

So boat must have an opposite force on front end with magnitude 45 N

Now similar way we can say

from back side of the boat the boy jumps off with force 60 N which means as per Newton's III law if boy has a force of 60 N in backward direction then he must apply a reaction force on the boat in reverse direction of same magnitude

So boat must have an opposite force on front end with magnitude 60 N

So here net force due to both jump on the boat is given by

$F_{net} = F_1 - F_2$