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3 years ago

Which describes the best approach when conducting a scientific experiment?

Physics

1 answer:

Varvara68 [4.7K]3 years ago

3 0

Answer:

A conduct repeated trials and make sure the experiment can be replicated.

Explanation:

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Help? its due in like 12 minutes lolzz

Gnom [1K]

Answer:

Question 1: the plates are moving toward one another.

Question 2: The Himalayan Mountains in India

Question 3: Because mountains are formed instead.

Explanation:

The paragraph explains that the plates continue to move closer to one another while forming multiple mountains.

The paragraph explains, " a well-known example of this is the formation of the Himalayan Mountains in India,"

The area of the Himalayan Mountains are better suited for the formation of mountains rather than volcanoes.

Have a nice day!! Good Luck!! Brainliest would be appreciated!!!

4 0

3 years ago

Given a particle that has the velocity v(t) = 3 cos(mt) = 3 cos (0.5t) meters, a. Find the acceleration at 3 seconds. b. Find th

DiKsa [7]

Answer:

a. $\rm -1.49\ m/s^2.$

b. $\rm 50.49\ m.$

Explanation:

<u>Given:</u>

Velocity of the particle, v(t) = 3 cos(mt) = 3 cos (0.5t) .

The acceleration of the particle at a time is defined as the rate of change of velocity of the particle at that time.

$\rm a = \dfrac{dv}{dt}\\=\dfrac{d}{dt}(3\cos(0.5\ t ))\\=3(-0.5\sin(0.5\ t.))\\=-1.5\sin(0.5\ t).$

At time t = 3 seconds,

$\rm a=-1.5\sin(0.5\times 3)=-1.49\ m/s^2.$

<u>Note</u>:<em> The arguments of the sine is calculated in unit of radian and not in degree.</em>

The velocity of the particle at some is defined as the rate of change of the position of the particle.

$\rm v = \dfrac{dr}{dt}.\\\therefore dr = vdt\Rightarrow \int dr=\int v\ dt.$

For the time interval of 2 seconds,

$\rm \int\limits^2_0 dr=\int\limits^2_0 v\ dt\\r(t=2)-r(t=0)=\int\limits^2_0 3\cos(0.5\ t)\ dt$

The term of the left is the displacement of the particle in time interval of 2 seconds, therefore,

$\Delta r=3\ \left (\dfrac{\sin(0.5\ t)}{0.05} \right )\limits^2_0\\=3\ \left (\dfrac{\sin(0.5\times 2)-sin(0.5\times 0)}{0.05} \right )\\=3\ \left (\dfrac{\sin(1.0)}{0.05} \right )\\=50.49\ m.$

It is the displacement of the particle in 2 seconds.

7 0

4 years ago

The water flowing through a 2.0 cm (inside diameter) pipe flows out through three 1.3 cm pipes. (a) If the flow rates in the thr

Verdich [7]

Answer:

a)54L/min

b)0.845

Explanation:

a) A x V= $A_1V_1+ A_2V_2+A_3V_3$

where suffix 1,2,3 refers to the three pipes.

=27L/min+16L/min+11 L/min

=54L/min

b) A x V=54L/min => $\frac{\pi }{4} d^2$ x v

d= 2 cm

$\frac{\pi }{4} d^2$ x v = 54

v= $\frac{4}{\pi }$ x $\frac{54}{2^2}$

-> $A_1$ x $V_1$ =27L/min => $\frac{\pi }{4} d_1^2$ x $v_1$

$d_1$ = 1.3cm

$\frac{\pi }{4} d^2$ x $v_1$ = 27

$v_1$ = $\frac{4}{\pi }$ x $\frac{27}{1.3^2}$

Next is to find the ratio of speed i.e $\frac{v}{v_1}$

$\frac{4}{\pi }$ x $\frac{54}{2^2}$ / $\frac{4}{\pi }$ x $\frac{27}{1.3^2}$ => $\frac{54}{27}$ $\frac{1.3^2}{2^2}$

$\frac{v}{v_1}$ = 0.845

8 0

3 years ago

A 34-m length of wire is stretched horizontally between two vertical posts. The wire carries a current of 68 A and experiences a

il63 [147K]

Answer:

7.28×10⁻⁵ T

Explanation:

Applying,

F = BILsin∅............. Equation 1

Where F = magnetic force, B = earth's magnetic field, I = current flowing through the wire, L = Length of the wire, ∅ = angle between the field and the wire.

make B the subject of the equation

B = F/ILsin∅.................. Equation 2

From the question,

Given: F = 0.16 N, I = 68 A, L = 34 m, ∅ = 72°

Substitute these values into equation 2

B = 0.16/(68×34×sin72°)

B = 0.16/(68×34×0.95)

B = 0.16/2196.4

B = 7.28×10⁻⁵ T

7 0

3 years ago

A catapult launches a test rocket vertically upward from a well, giving the rocket an initial speed of 80.6 m/s at ground level.

galina1969 [7]

Answer:

44.64 seconds

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration due to gravity = 9.8 m/s²

$v^2-u^2=2as\\\Rightarrow v=\sqrt{2as+u^2}\\\Rightarrow v=\sqrt{2\times 4.2\times 1180+80.6^2}\\\Rightarrow v=128.01\ m/s$

$v=u+at\\\Rightarrow 128.01=80.6+4.2t\\\Rightarrow t=\frac{128.01-80.6}{4.2}=11.29\ s$

<u>Time taken to reach 1180 m is 11.29 seconds</u>

$v=u+at\\\Rightarrow 0=128.01-9.8t\\\Rightarrow t=\frac{128.01}{9.8}=13.06\ s$

<u>Time the rocket will keep going up after the engines shut off is 13.06 seconds.</u>

$v^2-u^2=2as\\\Rightarrow s=\frac{v^2-u^2}{2a}\\\Rightarrow s=\frac{0^2-128.01^2}{2\times -9.8}\\\Rightarrow s=836.05\ m$

The distance the rocket will keep going up after the engines shut off is 836.05 m

Total distance traveled by the rocket in the upward direction is 1180+836.05 = 2016.05 m

The rocket will fall from this height

$s=ut+\frac{1}{2}at^2\\\Rightarrow 2016.05=0t+\frac{1}{2}\times 9.8\times t^2\\\Rightarrow t=\sqrt{\frac{2016.05\times 2}{9.8}}\\\Rightarrow t=20.29\ s$

<u>Time taken by the rocket to fall from maximum height is 20.29 seconds</u>

Time the rocket will stay in the air is 11.29+13.06+20.29 = 44.64 seconds

5 0

3 years ago