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3 years ago

You push against a wall and the wall pushes back on you.

Physics

1 answer:

emmainna [20.7K]3 years ago

4 0

Answer:

Its the third law of motion.

Explanation:

Newton's third law of motion states that every action has equal and opposite reaction.

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Which of the following in NOT an ionic compound? NaBr MgCl2 O SCI2 K (NO3)

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2 years ago

3) A tolley of mass 4kg, moving with a velocity

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Answer:

3.57 m/s

Explanation:

The sum of the 2 momentums Is equal the finale momentums. so if momentums Is q, v Is velocity and m Is Mass, q3=m1*v1+m2**v2=16+9=25 m*kg/s

q3=m3*v3

v3=q3/m3=25/(4+3)=3.57m/s

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2 years ago

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A 90 kg ice skater moving at 12.0 m/s on the ice encounters a region of roughed up ice with a coefficient of kinetic friction of

balandron [24]

Answer:

The skater covers a distance of 15 m before stopping.

Explanation:

Let the distance traveled before stopping be 'd' m.

Given:

Mass of the skater (m) = 90 kg

Initial velocity of the skater (u) = 12.0 m/s

Final velocity of the skater (v) = 0 m/s (Stops finally)

Coefficient of kinetic friction (μ) = 0.490

Acceleration due to gravity (g) = 9.8 m/s²

Now, we know that, from work-energy theorem, the work done by the net force on a body is equal to the change in its kinetic energy.

Here, the net force acting on the skater is only frictional force which acts in the direction opposite to motion.

Frictional force is given as:

$f=\mu N$

Where, 'N' is the normal force acting on the skater. As there is no vertical motion, $N=mg$

∴ $f=\mu mg=0.490\times 90\times 9.8=432.18\ N$

Now, work done by friction is a negative work as friction and displacement are in opposite direction and is given as:

$W=-fd=-432.18d$

Now, change in kinetic energy is given as:

$\Delta K=\frac{1}{2}m(v^2-u^2)\\\\\Delta K=\frac{1}{2}\times 90(0-12^2)\\\\\Delta K=45\times (-144)=-6480\ J$

Therefore, from work-energy theorem,

$W=\Delta K\\\\-432.18d=6480\\\\d=\frac{6480}{432.18}\\\\d=14.99\approx 15\ m$

Hence, the skater covers a distance of 15 m before stopping.

7 0

3 years ago

A projectile is fired with an initial velocity of 120.0 meters per second at an angle, θ above the horizontal. If the projectile

k0ka [10]

Answer:

θ = 62.72°

Explanation:

The projectile describes a parabolic path:

The parabolic movement results from the composition of a uniform rectilinear motion (horizontal ) and a uniformly accelerated rectilinear motion of upward or downward motion (vertical ).

The equation of uniform rectilinear motion (horizontal ) for the x axis is :

x = x₀+ vx*t Formula (1)

vx = v₀x

Where:

x: horizontal position in meters (m)

x₀: initial horizontal position in meters (m)

t : time (s)

vx: horizontal velocity in m/s

v₀x: Initial speed in x in m/s

The equations of uniformly accelerated rectilinear motion of upward (vertical ) for the y axis are:

y= y₀+(v₀y)*t - (1/2)*g*t² Formula (2)

vfy= v₀y -gt Formula (3)

Where:

y: vertical position in meters (m)

y₀ : initial vertical position in meters (m)

t : time in seconds (s)

v₀y: initial vertical velocity in m/s

vfy: final vertical velocity in m/s

g: acceleration due to gravity in m/s²

Data

v₀ = 120 m/s , at an angle θ above the horizontal

v₀x= 55 m/s

x-y components of the initial velocity ( v₀)

v₀x = v₀*cosθ Equation (1)

v₀y = v₀*sinθ Equation (2)

Calculating of the angle θ

We replace data in the Equation (1)

55 = 120*cosθ

cosθ = 55 / 120

$\theta = cos^{-1}( \frac{55}{120} )$

θ = 62.72°

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2 years ago