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3 years ago

Grade 10 My smart Physics people help me with this review question please

Physics

1 answer:

Kaylis [27]3 years ago

3 0

Answer: a = 1, b = 9, c = 6, d = 4

Explanation:

With no velocity, total mechanical energy is all gravity potential energy

PE = mgh = 80.0(9.81)(25) = 19,620 J = 1.96e4 J

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Three observers watch a train pull away from a station toward the right of the platform. Observer A is in one of the train’s car

juin [17]

Observer A is moving inside the train

so here observer A will not be able to see the change in position of train as he is standing in the same reference frame

So here as per observer A the train will remain at rest and its not moving at all

Observer B is standing on the platform so here it is a stationary reference frame which is outside the moving body

So here observer B will see the actual motion of train which is moving in forward direction away from the platform

Observer C is inside other train which is moving in opposite direction on parallel track. So as per observer C the train is coming nearer to him at faster speed then the actual speed because they are moving in opposite direction

So the distance between them will decrease at faster rate

Now as per Newton's II law

F = ma

Now if train apply the brakes the net force on it will be opposite to its motion

So we can say

- F = ma

$a = \frac{-F}{m}$

so here acceleration negative will show that train will get slower and its distance with respect to us is now increasing with less rate

It is not affected by the gravity because the gravity will cause the weight of train and this weight is always counterbalanced by normal force on the train

So there is no effect on train motion

5 0

4 years ago

A motorcycle of mass 100 kilograms slowly rolls off the edge of a cliff and falls for three seconds before reaching the bottom o

Jet001 [13]

Answer:

C) 3,000 kg m/s

Explanation:

We can consider the horizontal velocity of the motorcycle to be zero, since it rolls off the edge of the cliff very slowly. So, we only need to find the vertical velocity at the time of the impact with the ground.

The vertical velocity of the motorcycle at time t is given by (free-fall motion):

$v(t)=v_0 -gt$

where

$v_0=0$ is the initial vertical velocity (zero, since the motorcycle is not moving)

g = 9.8 m/s^2 is the acceleration due to gravity

t is the time

Since the motorcycle hits the ground after t = 3 seconds, we have

$v(3 s)=0-(9.8 m/s^2)(3 s)=-29.4 m/s$

And since we know its mass, m=100 kg, we can find its momentum:

$p=mv=(100 kg)(-29.4 m/s)=-2940 kg m/s \cdot -3000 kg m/s$

and the negative sign simply means downward direction.

8 0

3 years ago

A 10.0-kg box starts at rest on a level floor. An external, horizontal force of 2.00 × 102 N is applied to the box for a distanc

Harman [31]

Answer:

vf = 11.2 m/s

Explanation:

m = 10 Kg

F = 2*10² N

x = 4.00 m

μ = 0.44

vi = 0 m/s

vf = ?

We can apply Newton's 2nd Law

∑ Fx = m*a (→)

F - Ffriction = m*a ⇒ F - (μ*N) = F - (μ*m*g) = m*a ⇒ a = (F - μ*m*g)/m

⇒ a = (2*10² N - 0.44*10 Kg*9.81 m/s²)/10 Kg = 15.6836 m/s²

then , we use the equation

vf² = vi² + 2*a*x ⇒ vf = √(vi² + 2*a*x)

⇒ vf = √((0)² + 2*(15.6836 m/s²)*(4.00m)) = 11.2 m/s

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3 years ago

You are less likely to see a total solar eclipse than a total lunar eclipse because a. the moon’s shadow covers all of Earth dur

Bad White [126]

D. Because the moons shadow during a total lunar eclipse is tinnier than the earth.

8 0

3 years ago

Read 2 more answers

A 5.30kg block hangs from a spring with a spring constant 1700 N/m. The block is pulled down 4.50cm from the equilibrium positio

Andru [333]

To solve this problem it is necessary to apply the concepts related to the frequency in a spring, the conservation of energy and the total mechanical energy in the body (kinetic or potential as the case may be)

PART A) By definition the frequency in a spring is given by the equation

$f = \frac{1}{2\pi} \sqrt{\frac{k}{m}}$

Where,

m = mass

k = spring constant

Our values are,

k=1700N/m

m=5.3 kg

Replacing,

$f = \frac{1}{2\pi} \sqrt{\frac{1700}{5.3}}$

$f=2.85 Hz$

PART B) To solve this section it is necessary to apply the concepts related to the conservation of energy both potential (simple harmonic) and kinetic in the spring.

$\frac{1}{2}kA^2 = \frac{1}{2}mv^2 + \frac{1}{2} kY^2$