1) Propose un protocole permettant de mesurer la masse de 10 mL de chaque

A ball is thrown vertically upwards with a velocity of 30m/s. Determine the maximum height reached

padilas [110]

The maximum height reached is 45.92 m

6 0

2 years ago

A space vehicle is traveling at 2980 km/h relative to Earth when the exhausted rocket motor is disengaged and sent backward. The

Strike441 [17]

Answer:

3054.4 km/h

Explanation:

Using the conservation of momentum

momentum before separation = 5M × 2980 Km/h where M represent the mass of the module while 4 M represent the mass of the motor

initial momentum = 14900 M km/h

let v be the new speed of the motor so that the

new momentum = 4Mv and the new momentum of the module = M ( v + 94 km/h )

total momentum = 4Mv + Mv + 93 M = 5 Mv + 93M

initial momentum = final momentum

14900 M km/h = 5 Mv + 93M

14900 km/h = 5v + 93

14900 - 93 = 5v

v = 2961.4 km/h

the speed of the module = 2961.4 + 93 = 3054.4 km/h

8 0

3 years ago

Having aced your Physics 2111 class, you get a sweet summer-job working in the International Space Station. Your room-mate, Cosm

Sphinxa [80]

Answer:

a

The speed is $s = 5.857 m/s$

b

The distance is $D = 22.4 \ m$

Explanation:

From the question we are told that

The speed of the banana is $v = 16 \ m/s$

The distance from my location is $d = 8.2 \ m$

The time taken is $t = 1.4 \ s$

The speed of the ice cream is

$s = \frac{d}{t}$

substituting values

$s = \frac{8.4}{1.4}$

$s = 5.857 m/s$

The distance of separation between i and Valdimir is the same as the distance covered by the banana

So

$D = v * t$

substituting values

$D = 16 * 1.4$

$D = 22.4 \ m$

3 0

3 years ago

A uniform meter rule of weight 0.9N is suspended horizontally by two vertical loops of thread A and B placed 20cm and 30cm from

Rzqust [24]

The distance from the centre of the rule at which a 2N weight must be suspend from A is 29.3 cm.

<h3>Distance from the center of the meter rule</h3>

The distance from the centre of the rule at which a 2N weight must be suspend from A is calculated as follows;

-----------------------------------------------------------------

20 A (30 - x)↓ x ↓ 20 cm B 30 cm

2N 0.9N

Let the center of the meter rule = 50 cm

take moment about the center;

2(30 - x) + 0.9(x)(30 - x) = 0.9(20)

(30 - x)(2 + 0.9x) = 18

60 + 27x - 2x - 0.9x² = 18

60 + 25x - 0.9x² = 18

0.9x² - 25x - 42 = 0

x = 29.3 cm

Thus, the distance from the centre of the rule at which a 2N weight must be suspend from A is 29.3 cm.

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7 0

2 years ago

Read 2 more answers

An astronaut on the moon drops a feather from rest (v = 0). The acceleration due to gravity is 1.67 m/s 2 . If the feather beg

olga_2 [115]

Given :

Initial velocity , u = 0 m/s .

Acceleration due to gravity on moon , $g_m=1.67\ m/s^2$ .

Height , h = 2 m .

To Find :

Final position after falling for 1.5 seconds .

Solution :

We know , by equation of motion :

$s=ut+\dfrac{at^2}{2}$

Here , $a = g_m$ .

So , equation will transform by :

$s=ut+\dfrac{g_mt^2}{2}\\\\s=0+\dfrac{1.67\times 1.5^2}{2}\ m\\\\s=1.88\ m$

Therefore , the height form moon's surface is 1.88 m .

Hence , this is the required solution .

6 0

3 years ago