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4 years ago

Picture isn’t the best but could use the help

Physics

1 answer:

Leona [35]4 years ago

8 0

Answer:

60 deg from Y-axis is 150 deg from X-axis

x = L cos theta = 23,2 cos 150 = -.867 * 23.2 = -20.1

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On being introduced to the laws of thermodynamics, a student retorted, “When the brakes of a moving car are applied, the kinetic

inn [45]

Answer:

No, not at all. This kinetic energy of car is converted/lost in the form of sound and heat under the tyre when driver apply brakes.

Explanation:

This law states that energy can neither be created nor destroyed but it changes from one form to the other.

In this case reduction of kinetic energy of car does not contradict the law but it changes into heat and sound energy. When driver apply brakes we hear the sound of Tyre due to friction with the road. The tyre become hot too.

So it is not in contradiction to the law of conservation of energy (First Law of Thermodynamics).

Please mark branliest if you are satisfied with the answer. Thanking you in anticipation.

5 0

3 years ago

Can some please help me

Brilliant_brown [7]

Answer:

Classification

– Drug (highly regulated)

§ A substance which changes body structure or function.

§ Stimulate hormone secretions

§ Looks like medicine and/or is administered differently than foods

– Dietary Supplement (not highly regulated)

§ Highly refined products

§ No positive nutritional value

Explanation:

https://web.cortland.edu/buckenmeyerp/Lecture14.html

3 0

3 years ago

Car A m= 2000kg<br> v=20 km/h <br> Car B m= 2000kg <br> v= 20km/h what is the location after impact

Darya [45]

Mass of car, M=2000kg

Velocity, V=72× 5/18 =20ms¹

Apply kinematic equation of motion

v² -u² = as

0 − 20² =2a×20

a=−10ms^-2

Breaking force, F=ma=2000×10=20 kN

Apply first kinematic equation

v = u +at

$t= \frac{v - u}{a} = \frac{0 - 20}{ - 10} = 2sec\\$

7 0

3 years ago

The driver of a car traveling 110 km/h slams on the brakes so that the car undergoes a constant acceleration, skidding to a comp

Ber [7]

Answer: $-6.80\ ms^{-2}$

Explanation:

We know that the formula for acceleration is given by:

$a=\dfrac{v-u}{t}$ , where v = Final velocity

u= Initial velocity

Given : The driver of a car traveling 110 km/h slams on the brakes so that the car undergoes a constant acceleration.

i.e. u= 110 km/h $=\dfrac{110\times1000}{3600}\approx35.6\ m/s$ [∵ 1 km= 100 meters and 1 hour = 3600 seconds]

v= 0 m/s ( At brake , final velocity becomes 0)

t=4.5 seconds

Substitute all the values in the formula , we get

$a=\dfrac{0-30.56}{4.5}\approx-6.80\ ms^{-2}$

Hence, the average acceleration of the car during braking is $-6.80\ ms^{-2}$ .

5 0

3 years ago

An elevator together with its passengers weights 5000 N. At a certain instant, the tension in its supporting cable is 6000 N. De

gayaneshka [121]

We have to forces acting on the system (elevator+passengers):
1) The weight (W=5000 N), acting downward
2) The cable's tension (T=6000 N), acting upward
So, the two forces have opposite direction. The resultant (in upward direction) will be
$F=T-W$
And for Newton's second law, the resultant of the forces acting on the system causes an acceleration on the system itself, given by
$a= \frac{F}{m}$
where m is the mass of the system.

So, we need to find F and m.
The resultant of the forces is
$F=T-W=6000 N-5000 N=1000 N$
To find m, we can use the weight of the system. In fact, the weight of an object is given by
$W=mg$
where $g=9.81 m/s^2$ . Solving for m, and using W=5000 N, we find
$m= \frac{W}{g}= \frac{5000 N}{9.81 m/s^2}=510 kg$

and at this point, we can calculate the acceleration of the system (elevator+people):
$a= \frac{F}{m}= \frac{1000 N}{510 kg}=1.96 m/s^2$
and the acceleration has the same direction of the resultant force, so upward.

8 0

3 years ago