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2 years ago

30 POINTS PLS HELP NO LINKS PLS :/

Physics

2 answers:

seraphim [82]2 years ago

4 0

Answer:

Explanation:

trust me

Tresset [83]2 years ago

3 0

It’s c the same thing that is the same

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During adverse weather conditions such as rain or fog, drivers should take action accordingly by turning on their headlights, sl

xeze [42]

During adverse weather conditions such as rain or fog, drivers should take action accordingly by turning on their headlights, slowing down and increasing following distance. Adverse weather means that you are driving in difficult and dangerous conditions. Increasing following distance will help you to maintain safe driving and avoid tailgating.

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2 years ago

What is the acceleration due to gravity? А B 9.2 12.3 С 8.83 D 9.81

SCORPION-xisa [38]

Correct answer:

D) 9.81

8 0

2 years ago

A crate slides down a ramp that makes a 20∘ angle with the ground. To keep the crate moving at a steady speed, Paige pushes back

VMariaS [17]

Answer:

Hence, work done= 287.54 J

Explanation:

Given data:

angle of ramp with the ground θ =20°

force applied = 76 N

work done on the crate to slide down 4 m down the ramp

W= F×d cosθ ( only the cos component of the force will slide the crate down)

W= 76×4×cos20= 287.54 J

4 0

3 years ago

A 2.45-kg frictionless block is attached to an ideal spring with force constant 355 N/m. Initially the spring is neither stretch

ANTONII [103]

Answer:

A. A = 0.913 m

B. amax = 132.24m/s^2

C. Fmax = 324.01N

Explanation:

When the block is moving at the equilibrium point , its velocity is maximum.

A. To find the amplitude of the motion you use the following formula for the maximum velocity:

$v_{max}=A\omega$ (1)

vmax = maximum velocity = 11.0 m/s

A: amplitude of the motion = ?

w: angular frequency = ?

Then, you have to calculate the angular frequency of the motion, by using the following formula:

$\omega=\sqrt{\frac{k}{m}}$ (2)

k: spring constant = 355 N/m

m: mass of the object = 2.54 kg

$\omega = \sqrt{\frac{355N/m}{2.45kg}}=12.03\frac{rad}{s}$

Next, you solve the equation (1) for A and replace the values of vmax and w:

$A=\frac{v}{\omega}=\frac{11.0m/s}{12.03rad/s}=0.913m$

The amplitude of the motion is 0.913m

B. The maximum acceleration of the block is given by:

$a_{max}=A\omega^2 = (0.913m)(12.03rad/s)^2=132.24\frac{m}{s^2}$

The maximum acceleration is 132.24 m/s^2

C. The maximum force is calculated by using the second Newton law and the maximum acceleration:

$F_{max}=ma_{max}=(2.45kg)(132.24m/s^2)=324.01N$

It is also possible to calculate the maximum force by using:

Fmax = k*A = (355N/m)(0.913m) = 324.01N

The maximum force exertedbu the spring on the object is 324.01 N

4 0

2 years ago

A 1870 kg car traveling at 13.5 m/s collides with a 2970 kg car that is initally at rest at a stoplight. The cars stick together

Inga [223]

Answer:

The value is $\mu = 0.72$

Explanation:

From the question we are told that

The mass of the first car is $m_1 = 1870\ kg$

the initial speed of the car is $u = 13.5 \ m/s$

The mass of the second car is $m_2 = 2970\ kg$

The distance move by both cars is s = 1.93 m

Generally from the law of momentum conservation

$m_1 * u_1 + m_2 * u_2 = (m_1 + m_2 ) * v_f$

Here $u_2 = 0$ because the second car is at rest

and $v_f$ is the final velocity of the the two car

$1870* 13.5+ 0= ( 1870 + 2970 ) * v_f$

=> $v_f = 5.22\ m/s$

Generally from kinematic equation

$v_f^2 = u_2^2 + 2as$

here a is the deceleration

$5.22^2 = 0 + 2 *a * 1.93$

=> $a = 7.06 \ m/s^2$

Generally the frictional force is equal to the force propelling the car , this can be mathematically represented as

$F_f = F$

Here F is mathematically represented as

$F = (m_1 + m_2) * a$

$F = (1870 + 2970) * 7.06$

$F =34170.4 \ N$

and

$F_f = \mu * (m_1 + m_2 ) * g$

$F_f = 47432 * \mu$

$47432 * \mu = 34170.4$

=> $47432 * \mu = 34170.4$

=> $\mu = 0.72$

6 0

2 years ago