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3 years ago

2 Grade 11 Physics Questions. Will Mark BRAINLIEST!!!

Physics

2 answers:

Stels [109]3 years ago

7 0

Answer:

For the first one it would be screw and for the second one it is a ramp

Explanation:

Hope this helps

Viktor [21]3 years ago

3 0

Thanks for the help

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A stone is thrown from a cliff and lands in the sea. Air resistance is negligible. Which statement is correct whilst the stone i

erik [133]

Since the stone is travelling, the displacement cannot be constant. This leaves the last 2 choices as possibilities. Now consider the acceleration that can act on the stone. Acceleration due to gravity is the only one. That means the horizontal component of the stone has to be constant since there is no acceleration in the horizontal direction.

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3 years ago

Which of the following is a physical property?

coldgirl [10]

Flammability, I think!

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3 years ago

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Seth and Diana were asked to be the leaders for a game of tag during recess at school. Seth and Diana can ensure that all intere

grandymaker [24]

Answer:

Explanation:

did this yesterday with my friend

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2 years ago

The carbon isotope 14C is used for carbon dating of archeological artifacts. 14C(mass 2.34×10−26kg) decays by the process known

Nookie1986 [14]

Answer:

2240.92365 m/s

Explanation:

$m_1$ = Mass of electron = $9.11\times 10^{−31}\ kg$

$v_1$ = Speed of electron = $5.7\times 10^7\ m/s$

$p_2$ = Neutrino has a momentum = $7.3\times 10^{-24}\ kg m/s$

M = total mass = $2.34\times 10^{-26}\ kg$

In the x axis as the momentum is conserved

$Mv_x=m_1v_1\\\Rightarrow v_x=\dfrac{m_1v_1}{M}\\\Rightarrow v_x=\dfrac{9.11\times 10^{−31}\times 5.7\times 10^7}{2.34\times 10^{-26}}\\\Rightarrow v_x=2219.10256\ m/s$

In the y axis

$Mv_y=p_2\\\Rightarrow v_y=\dfrac{p_2}{M}\\\Rightarrow v_y=\dfrac{7.3\times 10^{-24}}{2.34\times 10^{-26}}\\\Rightarrow v_y=311.96581\ m/s$

The resultant velocity is

$R=\sqrt{v_x^2+v_y^2}\\\Rightarrow R=\sqrt{2219.10256^2+311.96581^2}\\\Rightarrow R=2240.92365\ m/s$

The recoil speed of the nucleus is 2240.92365 m/s

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3 years ago

The height (in meters) of a projectile shot vertically upward from a point 2 m above ground level with an initial velocity of 22

alexgriva [62]

1) The law of motion of the projectile is
$h(t) = 2+22.5 t-4.9 t^2$
To find the velocity, we should compute the derivative of h(t):
$v(t)=h'(t)=22.5-2\cdot 4.9t=22.5-9.8t$
So now we can calculate the speed at t=2 s and t=4 s:
$v(2.0s)=22.5-9.8\cdot2.0 =2.9 m/s$
$v(4.0s)=22.5-9.8\cdot 4.0s=-16.7 m/s$
The negative sign in the second speed means the projectile has already reached its maximum height and it is now going downward.

2) The projectile reaches its maximum height when the speed is equal to zero:
$v(t)=0$
So we have
$22.5-9.8 t=0$
And solving this we find
$t=2.30 s$

3) To find the maximum height, we take h(t) and we just replace t with the time at which the projectile reaches the maximum height, i.e. t=2.30 s:
$h(2.30 s)=2+22.5\cdot 2.30 -4.9 \cdot (2.30s)^2 = 27.83 m$

4) The time at which the projectile hits the ground is the time at which the height is zero: h(t)=0. So, this translates into
$2+22.5t -4.9 t^2 = 0$
This is a second-order equation, and if we solve it we get two solutions: the first solution is negative, so we can ignore it since it's physically meaningless; the second solution is
$t=4.68 s$
And this is the time at which the projectile hits the ground.

5) The velocity of the projectile when it hits the ground is the velocity at time t=4.68 s:
$v(4.68 s)=22.5-9.8\cdot 4.68 =-23.36 m/s$
with negative sign, because it is directed downward.

8 0

4 years ago