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3 years ago

Please help.. urgent Which statement is equivalent to Newton's first law? a. 15,300 N b. 1.20*10^3 N c. 2,030 N d. 1,560 N

1 answer:

6 0

According to Newton laws of motion,
F = m*a
Here, m = 1,560 Kg
a = 1.30 m/s²

Substitute their values,
F = 1,560 * 1.30
F = 2028 N ~ 2030 N [ Closest value ]

In short, Your Answer would be Option C

Hope this helps!

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Which force changes the lithosphere by building up the surface?

Sunny_sXe [5.5K]

Answer:

volcanic eruptions

Explanation:

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3 years ago

A ladder of length 10 is leaning against a wall. The vertical wall is frictionless while the horizontal surface is rough. The an

bogdanovich [222]

Answer:

105.1N

Explanation:

According to the newton second law

$\sum F_x = ma_x\\F_m - F_f = ma_x\\$

Since the acceleration is zero, then;

$F_m - F_f = 0\\Fm = F_f\\$

Since;

$F_m = mg sin \theta, \ hence \ F_f = mg sin \theta$

Given the following

mass of ladder m = 14kg

acceleration due to gravity g = 9.8m/s²

θ = 50°

Substitute

$F_f = 14(9.8)sin50\\F_f = 14(7.5072)\\F_f = 105.1N\\$

Hence the magnitude of the friction force (in N) exerted on the ladder in the point of contact with the horizontal surface is 105.1N

8 0

3 years ago

The purpose of teaching high levels of math is to filter out students who lack logical intelligence

Pachacha [2.7K]

Answer: False I think that is worded weird tho

7 0

3 years ago

A spinning wheel having a mass of 20 kg and a diameter of 0.5 m is positioned to rotate about its vertical axis with a constant

AlekseyPX

Answer:

ωf = 8.8 rad/s

v = 2.2 m/s

Explanation:

We will use the third equation of motion to find the maximum angular velocity of the wheel:

$2\alpha \theta = \omega_f^2 -\omega_I^2$

where,

α = angular acceleration = 6 rad/s²

θ = angular displacemnt = 1 rev = 2π rad

ωf = max. final angular velocity = ?

ωi = initial angular velocity = 1.5 rad/s

Therefore,

$2(6\ rad/s^2)(2\pi\ rad)=\omega_f^2-(1.5\ rad/s)^2\\\omega_f^2=75.4\ rad/s^2+2.25\ rad/s^2\\\omega_f = \sqrt{77.65\ rad/s^2}$

Now, for linear velocity:

v = rω = (0.25 m)(8.8 rad/s)

7 0

3 years ago

A torque of 36.5 N · m is applied to an initially motionless wheel which rotates around a fixed axis. This torque is the result

vivado [14]

Answer:

$21.6\ \text{kg m}^2$

$3.672\ \text{Nm}$

$54.66\ \text{revolutions}$

Explanation:

$\tau$ = Torque = 36.5 Nm

$\omega_i$ = Initial angular velocity = 0

$\omega_f$ = Final angular velocity = 10.3 rad/s

t = Time = 6.1 s

I = Moment of inertia

From the kinematic equations of linear motion we have

$\omega_f=\omega_i+\alpha_1 t\\\Rightarrow \alpha_1=\dfrac{\omega_f-\omega_i}{t}\\\Rightarrow \alpha_1=\dfrac{10.3-0}{6.1}\\\Rightarrow \alpha_1=1.69\ \text{rad/s}^2$

Torque is given by

$\tau=I\alpha_1\\\Rightarrow I=\dfrac{\tau}{\alpha_1}\\\Rightarrow I=\dfrac{36.5}{1.69}\\\Rightarrow I=21.6\ \text{kg m}^2$

The wheel's moment of inertia is $21.6\ \text{kg m}^2$

t = 60.6 s

$\omega_i$ = 10.3 rad/s

$\omega_f$ = 0

$\alpha_2=\dfrac{0-10.3}{60.6}\\\Rightarrow \alpha_1=-0.17\ \text{rad/s}^2$

Frictional torque is given by

$\tau_f=I\alpha_2\\\Rightarrow \tau_f=21.6\times -0.17\\\Rightarrow \tau=-3.672\ \text{Nm}$

The magnitude of the torque caused by friction is $3.672\ \text{Nm}$

Speeding up

$\theta_1=0\times t+\dfrac{1}{2}\times 1.69\times 6.1^2\\\Rightarrow \theta_1=31.44\ \text{rad}$

Slowing down

$\theta_2=10.3\times 60.6+\dfrac{1}{2}\times (-0.17)\times 60.6^2\\\Rightarrow \theta_2=312.03\ \text{rad}$

Total number of revolutions

$\theta=\theta_1+\theta_2\\\Rightarrow \theta=31.44+312.03=343.47\ \text{rad}$

$\dfrac{343.47}{2\pi}=54.66\ \text{revolutions}$

The total number of revolutions the wheel goes through is $54.66\ \text{revolutions}$ .

3 0

3 years ago