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3 years ago

Hardness and density are physical properties. a. True b. False

Physics

2 answers:

antiseptic1488 [7]3 years ago

7 0

a . true hardness and density are physical properties

VMariaS [17]3 years ago

5 0

Answer: The Answer would be True.

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Y=1/2 at^2 solve for t

Aloiza [94]

Explanation:

y = ½ at²

Multiply both sides by 2:

2y = at²

Divide both sides by a:

t² = 2y/a

Take the square root of both sides:

t = √(2y/a)

6 0

3 years ago

If a ball is 10m high with what velocity will it fall?

Semmy [17]

14m/s

Explanation:

Given parameters:

Height of the ball = 10m

Unknown:

Velocity of fall or final velocity = ?

Solution:

We are going to use the appropriate equation of motion to solve this problem.

The object is falling with respect to gravity.

V² = U² + 2gH

where V is the final velocity

U is the initial velocity

g is the acceleration due to gravity 9.8m/s²

H is the height of fall

The initial velocity here is zero and

V² = 2 x 9.8 x 10 = 196

V = 14m/s

learn more:

Motion problems brainly.com/question/5248528

#learnwithBrainly

6 0

3 years ago

If you walk at an average speed of 5 km/h for 30 minutes, how

timurjin [86]

Time= s/v

Speed =5km/h
Time=30min
Distance is required
Distance=time*speed
30min*5km/h=600m

4 0

3 years ago

The electron beam inside a television picture tube is 0.40 {\rm mm} in diameter and carries a current of 50 {\rm \mu A}. This el

Harrizon [31]

Answer:

A.3.13x10^14 electrons

B.330A/m²

C.9.11x10^5N/C

D. 0.23W

.pls see attached file for explanations

8 0

3 years ago

A 75-g bullet is fired from a rifle having a barrel 0.540 m long. choose the origin to be at the location where the bullet begin

lyudmila [28]

Part a) The work done by the gas on the bullet is the integral of the force in dx, where x is the distance covered by the bullet inside the barrel with respect to the origin:
$W= \int\limits^{0.540m}_{0} {F} \, dx = \int\limits^{0.540m}_{0} {(16000+10000x-26000x^2)} \, dx =$
$=16000x+10000 \frac{x^2}{2} - 26000 \frac{x^3}{3}$
By substituting the length of the barrel, L=0.540 m, we find the total work done by the gas on the bullet:
$W=16000(0.540m)+10000 \frac{(0.540m)^2}{2} - 26000 \frac{(0.540m)^3}{3} =$
$=8733 J=8.73 kJ$

part b) The resolution of the problem is the same, we just have to use the new length of the barrel (L=0.95 m) inside the final formula, and we find the new value of the work:
$W=16000(0.95m)+10000 \frac{(0.95m)^2}{2} - 26000 \frac{(0.95m)^3}{3} =$
$=12280 J=12.28 kJ$

5 0

3 years ago