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2 years ago

A 4 kg block has a coefficient of friction of 0.3 between itself and the surface

Physics

1 answer:

Deffense [45]2 years ago

4 0

Answer:

C - 11.76 N (Newtons)

Explanation:

2 steps are needed,

First find the Force being applied by the box on the surface.

F = M(a) [Force equals: mass X Acceleration (of gravity)]

You get 39.2 N of Force.

Plug in the remaining values into the equation for Frictional Force

F(f) = μ(F) [Coefficient of Friction X Force applied by the Box]

F(f) = 0.3(39.2) = 11.76 N (of Frictional Force)

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Two children are throwing a ball back-and-forth straight across the back seat of a car. The ball is being thrown 7 mph relative

Evgesh-ka [11]

Answer:

the ball will fly in AX direction, making angle of 8.84° from the motion of the car

Explanation:

Given the data in the question and as illustrated in the diagram below;

Now, Lets assume line AB represent the movement of the car,

AC is the movement of the ball been thrown back and forth in the back seat

Ax is the motion of the ball it flies off the window

so from the diagram, We can see triangle ABC

where AB is 45 mph and AC = 7 mph

and angle ∠CAB = 90°

using SOH CAH TOA

TOA; tanθ = Opposite / Adjacent

tanθ = Opposite / Adjacent

tan( ∠ ABC ) = AC / AB

we substitute

tan( ∠ ABC ) = 7 / 45

tan( ∠ ABC ) = 0.15555

( ∠ ABC ) = tan⁻¹ 0.15555

( ∠ ABC ) = 8.84°

Therefor, angle ( ∠ ABC ) is 8.84°

Meaning angle ( ∠ XAA' ) is also 8.84°

Therefore, the ball will fly in AX direction, making angle of 8.84° from the motion of the car

4 0

2 years ago

Mike is standing on the roof of a building looking at the roof of the neighboring building that is 15 meters away and 10 meters

amid [387]

Answer:

Part a)

$t = 1.65 s$

Part b)

$x = 40.4 m$

Since the distance of other building is 15 m so YES it can make it to other building

Part c)

$v = 27.3 m/s$

direction of velocity is given as

$[tex]\theta = 26.35 degree$

Explanation:

Part a)

acceleration due to gravity on this planet is 3/4 times the gravity on earth

So the acceleration due to gravity on this new planet is given as

$a = \frac{3}{4}(9.81)$

$a = 7.36 m/s^2$

now the vertical displacement covered by the canister is given as

$y = 10 m$

now by kinematics we have

$y = \frac{1}{2}gt^2$

$10 = \frac{1}{2}(7.36)t^2$

$t = 1.65 s$

Part b)

Horizontal speed of the canister is given as

$v_x = 24.5 m/s$

now the distance moved by it

$x = v_x t$

$x = 24.5 (1.65)$

$x = 40.4 m$

Since the distance of other building is 15 m so YES it can make it to other building

Part c)

Final velocity in X direction will remains the same

$v_x = 24.5 m/s$

final velocity in Y direction

$v_y = v_i + at$

$v_y = 0 + (7.36)(1.65)$

$v_y = 12.14 m/s$

now magnitude of velocity is given as

$v = \sqrt{v_x^2 + v_y^2}$

$v = \sqrt{24.5^2 + 12.14^2}$

$v = 27.3 m/s$

direction of velocity is given as

$\theta = tan^{-1}\frac{v_y}{v_x}$

$\theta = tan^{-1}\frac{12.14}{24.5}$

$[tex]\theta = 26.35 degree$

6 0

3 years ago

In the lab downstairs physics majors use a rotating mirror to measure the speed of light within a few percent of the actual valu

iris [78.8K]

The number of complete cycles the rotating mirror goes through before the angular velocity gets to zero is approximately 1166.8 revs

<h3>What is angular velocity?</h3>

Angular velocity is the ratio of the angle turned to the time taken.

The kinematic equation for angular velocity are presented as follows;

ω = ω₀ + α·t

θ = θ₀ + ω₀·t + 0.5·α·t²

Where;

θ₀ = The initial angle turned = 0

ω₀ = The initial angular velocity of the mirrors = 115 rad/s clockwise

α = The angular acceleration = (115 - (-115))rad/s/(85 s) = -46/17 m/s²

t = The duration of the motion;

When the angular velocity, ω is zero, we get;

0 = 115 - 46/17·t

t = 85/2

Which indicates;

θ = 0 + 115× (85/2) + 0.5×(46/17) ×(85/2)² = 7331.25

θ = 7331.25 radians

θ = 7331.25/(2×π) ≈ 1166.8 rev

The mirrors would have turned through approximately 1166.8 revolutions when the angular gets to zero

Learn more about angular velocity and acceleration here:

brainly.com/question/13014974

#SPJ1

7 0

1 year ago

A mass of 13kg stretches a spring 14cm. The mass is acted on by an external force of 6sin(t/2)N and moves in a medium that impar

m_a_m_a [10]

Answer:

$13u\prime\prime+33.33u\prime+910u=6sin\frac{t}{2}, \ u(0)=0,u\prime(0)=0.04\\$

#Where $u$ is in meters and $t$ in seconds.

Explanation:

Given that : $m=13.0kg, \ L=0.14m, \ F(t)=6sin\frac{t}{2}N, F_d(t*)=-4N^{-1}, u\prime(t*)=0.12m/s\\u(0)=0,u\prime(0)=0.04m/s$

From $\omega=kL$ we have:

$k=\frac{\omega}{L}=\frac{mg}{0.14m}=\frac{13.0\times 9.8m/s}{0.14m}\\\\k=910kg/s^2$

From $F_d(t)=-\gamma u\prime(t)$ we have that:

$\gamma=-\frac{F_d(t*)}{u\prime(t*)}=\frac{4N}{0.12m/s}\\=33.33Ns/m$

Now,given that the initial value problem is given by:

$13u\prime\prime+33.33u\prime+910u=6sin\frac{t}{2}, \ u(0)=0,u\prime(0)=0.04\\$

Hence,the position of u at time t is given by

$13u\prime\prime+33.33u\prime+910u=6sin\frac{t}{2}, \ u(0)=0,u\prime(0)=0.04\\$ , $u$ in meters, $t$ in seconds.

4 0

3 years ago

an optician uses a plane mirror to help him. suppse a patient sits in a chair 2.5m away from him. He views the image of a chart

viva [34]

Answer:

I think 75 m

Explanation:

tell if it was correct

5 0

3 years ago