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2 years ago

During a test, a solution placed on a sheet

Physics

1 answer:

suter [353]2 years ago

3 0

Answer:

B. Acidic

Explanation:

Neutral an Basic solutions wouldn't create any reaction

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A tennis player receives a shot with the ball (0.0600 kg) traveling horizontally at 22.0 m/s and returns the shot with the ball

morpeh [17]

I need an answer choice

8 0

2 years ago

Consider a compact car that is being driven

aliina [53]

Answer:

Height h = 37.8 m

Explanation:

Given :

Velocity of car (v) = 98 km / h

Acceleration of gravity = 9.8 m/s²

Computation:

Acceleration of gravity = 9.8 m/s²

Acceleration of gravity = (98)(1,000 m / 3,600 s)

Acceleration of gravity = 27.22 m/s

By using law of conservation of energy ;

(1/2)mv² = mgh

h = v² / 2g

h = 27.22² / 2(9.8)

Height h = 37.8 m

5 0

3 years ago

A box of weight 280 N is being pulled to the right by a pulling force vec F of magnitude 50 N. The box is moving at a constant s

boyakko [2]

Answer:

what diagram. I can't see any

5 0

2 years ago

Starting from rest, a 5.00-kg block slides 2.50 m down a rough 30.0 incline. The coeffi cient of kinetic friction between the bl

UkoKoshka [18]

Answer:

a) $W_g =61.25J$

b) $W_k = -46.25J$

c) $W_N = 0$

d) $W_g$ would be the same.

$W_k$ would decrease.

$W_N$ would be the same.

Explanation:

a) On an inclined plane the force of gravity is the sine component of the weight of the block.

$F_g = mg\sin(\theta) = 5(9.8)\sin(30^\circ)\\W_g = F_g x = 5(9.8)\sin(30^\circ)2.5 = 61.25J$

b) The friction force is equal to the normal force times coefficient of friction.

$F_k = -mg\cos(\theta)\mu_k = -5(9.8)cos(30^\circ)0.436 = -18.5 N\\W_k = -F_kx = -46.25J$

c) The work done by the normal force is zero, since there is no motion in the direction of the normal force.

d) The relation between the vertical height and the distance on the ramp is

$h = x\sin(\theta)$

According to this relation, the work done by the gravity wouldn't change, since the force of gravity includes a term of $x\sin(\theta)$ .

The work done by the friction force would decrease, because both the cosine term and the distance on the ramp would decline.

The work done by the normal force would still be zero.

5 0

2 years ago

How much heat would be needed to completely evaporate 31.5 g of boiling water at a temperature of 100 "C? Express your answer in

BaLLatris [955]

Answer:

Heat needed = 71.19 J

Explanation:

Here heat required can be calculated by the formula

H = mL

M is the mass of water and L is the latent heat of vaporization.

Mass of water, m = 31.5 g = 0.0315 kg

Latent heat of vaporization of water = 2260 kJ/kg

Substituting

H = mL = 0.0315 x 2260 = 71.19 kJ

Heat needed = 71.19 J

7 0

2 years ago