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ipn [44]
4 years ago
8

HELP PLEASE

Physics
2 answers:
Leto [7]4 years ago
8 0

Answer:

2.5m/s^2

Explanation:

We have the first equation of motion here:

V(f) = V(i) + at (equation 1)

Now,

V(f) = Final velocity = 25 m/s

V(i) = Initial velocity = 0 m/s (Since car starts moving from rest, so intial velocity has to be considered as 0 m/s)

a = Acceleration = ? (It has to be determined)

t= Time = 10 seconds

Now by putting values in equation 1

25 = 0 + (a)*(10)

25 = 10*a

a = 25/10

a = 2.5 m/s^2

Additionally, remember that we have three equations of motion

First equation is V(f) = V(i) + at

Second is S = V(i)t + (1/2)a(t^2)

Third is 2as = [V(f)]^2 - [V(i)]^2

Where S = Distance

            t = Time

            a= Acceleration

            V(f) = Final velocity

            V(i) = Initial velocity

We always decide to use any of these equations according to the data given, like you can see that I used first equation because it is independent of Distance, since in data, there was no distance given, so i used first equation of motion.

4vir4ik [10]4 years ago
7 0

Recall this kinematic equation:

a = \frac{Vi+Vf}{Δt}

This equation gives the acceleration of the object assuming it IS constant (the velocity changes at a uniform rate).

a is the acceleration.

Vi is the initial velocity.

Vf is the final velocity.

Δt is the amount of elapsed time.


Given values:

Vi = 0 m/s (the car starts at rest).

Vf = 25 m/s.

Δt = 10 s


Substitute the terms in the equation with the given values and solve for a:

a = \frac{0+25}{10}

<h3>a = 2.5 m/s²</h3>
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emmainna [20.7K]

Answer:

x-component of force is  38.18 lb where as magnitude of Force is 93.16

Explanation:

Fy of the force F exerted on the handle of the box wrench = 86 lb

Considering the triangle in Fig 1

magnitude of perpendicular = P =  12

magnitude of base = B = 5

using Pythagoras theorem

                        H= \sqrt{P^{2} + B^{2}}

                 H= \sqrt{12^{2} + 5^{2}}

            H=13\\\implies cos \theta = \frac{5}{13} \\\implies sin \theta = \frac{12}{13}\\

y-component of force is given given as:

                              86 = Fsin\theta\\F = 86 (\frac{13}{12})\\F = 93.16 lb\\F_{x} =F cos \theta\\F_{x} = 93.16 (\frac{5}{13})\\F_{x} =38.18 lb.

5 0
3 years ago
A sphere of mass of 1.55 kg is accelerated upwards by a string to which the sphere is attached. Its speed increases from 2.81 m/
Mama L [17]

Answer:

The tension in the string is 16.24 N

Explanation:

Given;

mass of the sphere, m = 1.55 kg

initial velocity of the sphere, u = 2.81 m/s

final velocity of the sphere, v = 4.60 m/s

duration of change in the velocity, Δt = 2.64 s

The tension of the string is calculated as follows;

T =  ma  + mg\\\\T = m(a + g)\\\\where;\\\\a \ is \ upward \ acceleration \ of \ the \ sphere\\\\g \ is \ acceleration \ due \ to \ gravity =9.8 \ m/s^2\\\\a = \frac{\Delta V}{\Delta t} = \frac{v- u}{ t} = \frac{4.6 - 2.81 }{2.64} = 0.678 \ m/s^2

T = 1.55(0.678 + 9.8)

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T = 16.24 N

Therefore, the tension in the string is 16.24 N

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3 years ago
if the magnitudes of the forces vary with time as F1=Ct and F = 2Ct, where C equals to 7.5 N/s and t is time, find the time t0 a
Degger [83]

Answer:

The tension in the string is equal to Ct

And the time t0 when the rension in the string is 27N is 3.6s.

Explanation:

An approach to solving this problem jnvolves looking at the whole system as one body by drawing an imaginary box around both bodies and taking summation of the forces. This gives F2 - F1 = Ct. This is only possible assuming the string is massless and does not stretch, that way transmitting the force applied across it undiminished.

So T = Ct

When T = 27N then t = T/C = 27/7.5 = 3.6s

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D because u r using your energy to do all the things

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3 years ago
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