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2 years ago

Which example is a simple machine?

Physics

1 answer:

kirza4 [7]2 years ago

7 0

Answer:

Ramp

Explanation:

All of the other things are fairly complex, bikes have to be made with multiple parts so do cars and steering wheels are another part of a boat, a ramp usually consists of two or fewer parts making it the simplest.

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what is the mass of a cannon ball if a force 2500 N gives the cannon ball an acceleration of 200 m/s squared?

nignag [31]

Using Newton's second law of motion:

F=ma ; [ F = force (N: kgm/s^2);m= mass (kg); a = acceleration (m/s^2)

Given: Find: Formula: Solve for m:

F: 2500N mass:? F=ma Eq.1 m=F/a Eq. 2

a= 200m/s^2

Solution:

Using Eq.2

m= (2500 kgm/s^2)/ (200m/s^2) = 12.5 kg

8 0

3 years ago

A rotating merry go-round makes one complete revolution in 4.2 s. (a) what is the linear speed of a child seated 1.3 m from the

gregori [183]

The radius, r, of the child from the center of the wheel is
r = 1.3 m

The wheel makes one revolution in 4.2 s. Its angular velocity is
ω = (2π rad)/(4.2 s) = 1.496 rad/s

The linear speed of the child is the tangential velocity, given by
v = rω
= (1.3 m)*(1.496 rad/s)
= 1.945 m/s

Answer: 1.95 m/s (nearest hundredth)

6 0

3 years ago

Read 2 more answers

Salmon often jump waterfalls to reach their breeding grounds. Starting downstream, 3.04 m away from a waterfall 0.585 m in heigh

S_A_V [24]

Answer:

V₀ = 5.47 m/s

Explanation:

The jumping motion of the Salmon can be modelled as the projectile motion. So, we use the formula for the range of projectile motion here:

R = V₀² Sin 2θ/g

where,

R = Range of Projectile = 3.04 m

θ = Launch Angle = 41.7°

V₀ = Minimum Launch Speed = ?

g = 9.81 m/s²

Therefore,

3.04 m = V₀² [Sin2(41.7°)]/(9.81 m/s²)

V₀² = 3.04 m/(0.10126 s²/m)

V₀ = √30.02 m²/s²

V₀ = 5.47 m/s

6 0

3 years ago

Two football players collide head-on in midair while trying to catch a thrown football. The first player is 98.5 kg and has an i

romanna [79]

Answer:

$v_f=0.825m/s$

Explanation:

We must use conservation of linear momentum before and after the collision, $p_i=p_f$

Before the collision we have:

$p_i=p_1+p_2=m_1v_1+m_2v_2$

where these are the masses are initial velocities of both players.

After the collision we have:

$p_f=(m_1+m_2)v_f$

since they clong together, acting as one body.

This means we have:

$m_1v_1+m_2v_2=(m_1+m_2)v_f$

Or:

$v_f=\frac{m_1v_1+m_2v_2}{m_1+m_2}$

Which for our values is:

$v_f=\frac{(98.5kg)(6.05m/s)+(119kg)(-3.5m/s)}{(98.5kg)+(119kg)}=0.825m/s$

7 0

3 years ago

Two objects must be in contact for them to exert a force on each other. True False

Irina-Kira [14]

Answer:

False

Explanation:

The given statement "Two objects must be in contact for them to exert a force on each other" is not true as there are many types of forces that doesn't require being in contact for exerting a force.

One such example is the gravitational force acting between two bodies. Gravitational force is the force of pull with which a body pulls another body without being in contact.

For two bodies of masses 'M' and 'm' separated at a distance of 'R', the gravitational force is given as:

$Force=\frac{GMm}{R^2}\\Where,G\to \textrm{Universal Gravitational constant}$

The gravitational force acts always act between bodies that have mass. The bodies are not in contact yet experience force.

Therefore, the given statement is false.

5 0

4 years ago