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4 years ago

In the figure a block slides along a track from one level to a higher level, by moving through an

Physics

1 answer:

Scrat [10]4 years ago

7 0

Answer:

(B) 1.17 m.

Explanation:

At the highest point, the conservation of energy equation would becomes;

K.E (initial ) = K.E + P.E;

½ m(V)^2 = ½ m(V)^2 + mgh

(m*6^2)/2 = (m * 9.81 * 1.1) + (m*V^2)/2

36/2 = (9.81 * 1.1) + v^2/2

Solving the equation;

v=√14.44 m/s

= 3.8 m/s

a = −0.6g

= -0.6*9.81

= −5.88 m/s^2

Using equation of motion,

V^2 = U^2 + 2aS

S = V^2/2a

= 14.44/12

= 1.19 m.

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A college dorm room measures 14 ft wide by 13 ft long by 6 ft high. What is the air in it under normal conditions?

kirza4 [7]

Complete question:

A college dormitory room measures 14 ft wide by 13 ft long by 6 ft high. Weight density of air is 0.07 lbs/ft3. What is the weight of air in it under normal conditions?

Answer:

the weight of the air is 76.44 lbs

Explanation:

Given;

dimension of the dormitory, = 14 ft by 13 ft by 6 ft

density of the air, = 0.07 lbs/ft³

The volume of the air in the dormitory room = 14 ft x 13 ft x 6 ft

= 1092 ft³

The weight of the air = density x volume

= 0.07 lbs/ft³ x 1092 ft³

= 76.44 lbs

Therefore, the weight of the air is 76.44 lbs

6 0

3 years ago

A rifle is aimed horizontally at a target 49 m away. the bullet hits the target 2.3 cm below the aim point. part a what was the

Nezavi [6.7K]

We use the equation of motion for vertical component,

$s_{y} = u_{y} t+\frac{1}{2} gt^2.$

Here, $s_{y}$ is displacement of bullet, $u_{y}$ is vertical initial velocity of bullet which is equal to zero because bullet was fired horizontally, and t is time of flight.

Therefore,

$s_{y} =\frac{1}{2}gt^2$

Given, $s_{y} =2.3 \ cm = 2.3 \times 10^{-2} \ m$

Substituting the values, we get time of flight

$2.3 \times 10^{-2} \ m = \frac{1}{2} \times 9.8 \ m/s^2 \times t^2 \\\\ t =\sqrt{46.94 \times 10^{-4} \ s } = 0.069 \ s$

4 0

3 years ago

A 5.00 gram bullet moving at 600 m/s penetrates a tree trunk to a depth of 4.00 cm.

Crazy boy [7]

A) Work energy relation;
Work =ΔKE ; work done = Force × distance, while, Kinetic energy = 1/2 MV²
F.s = 1/2mv²
F× 4×10^-2 = 1/2 × 5 ×10^-3 × (600)²
F = 900/0.04
= 22500 N
Therefore, force is 22500 N

b) From newton's second law of motion;
F = Ma
Thus; a = F/m
= 22500/(5×10^-3)
= 4,500,000 m/s²
But v = u-at
0 = 600- 4500,000 t
t = 1.33 × 10^-4 seconds

6 0

4 years ago

How does the hydraulic garage lift helps to lift heavy weights?Explain in you words don't copy (write in the easiest way you cou

Sergeeva-Olga [200]

Answer:

this the way that hydraulic pump works:

The system of hydraulic jack lifts the heavy objects by transferring the oil from one chamber to another.

Through the suction valves into the two cylinders which creates pressure distributing the fluid equally to the points causing the objects to lift up.

hope these points helped u

any confusion then comment it

5 0

3 years ago

Find the electric field at a point midway between two charges of 30.0×10 power -9 and 60.0×10 power -9 separated by a distance o

KATRIN_1 [288]

Answer:

The electric field at a point midway between the two charges, E = -1.8 * 10⁴ N/C

Explanation:

Let the midpoint of the two charges be considered as the origin, and charge A = 30.0 * 10⁻⁹ C be moving in the +x- axis and the charge B = 60.0 * 10⁻⁹ C be moving in the -x-axis.

Electric field, E = kQ/r² where k is a constant = 9.0 * 10⁹ N.m²/C², Q = quantity of charge, r = distance of separation

In the given question,r = 30.0 cm = 0.03 m; the midway point between A and B = 0.03/2 = 0.015 m

Electric field due to charge A

Ea = +(9.0 * 10⁹ N.m²/C² * 30.0 * 10⁻⁹ ) / ( 0.015 m)²

Ea = +1.8 * 10⁴ N/C

Electric field due to charge B

Eb = -(9.0 * 10⁹ N.m²/C² * 60.0 * 10⁻⁹ ) / ( 0.015 m)²

Eb = -3.6 * 10⁴ N/C

The resultant electric field E = Ea + Eb

E = (+1.8 * 10⁴ + -3.6 * 10⁴) N/C

E = -1.8 * 10⁴ N/C

Therefore, the electric field at a point midway between the two charges, E = -1.8 * 10⁴ N/C

7 0

3 years ago