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Roman55 [17]
2 years ago
8

A force of 19 newtons is applied on a cart of 2 kilograms, and it experiences a frictional force of 1.7 newtons. What is the acc

eleration of the cart
Physics
2 answers:
Dmitriy789 [7]2 years ago
5 0

Answer:

8.6 meters/second^2

Explanation:

Net force = 19 N - 1.7 N = 17.3 N

F = ma

17.3 N = 2 kg * a

a = 17.3 N / 2 kg  (N= kg*m/s^2)

a = 8.6 m/s^2

Umnica [9.8K]2 years ago
4 0
Frictional force always opposes applied force, so the net force on the cart would have to be 19N - 1.7N. The acceleration can then be solved by using the relation: F = ma. This is shown below:

Net force = 19 - 1.7 = 17.3 N

Acceleration = Force / mass
Acceleration = 17.3 / 2
Acceleration = 8.65 N/m
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At what angle are the electronic and the magnetic wave related in an electromagnetic signal?
VikaD [51]

Answer:

90degrees I'm pretty sure

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2 years ago
How can a force change the motion of an object that is already moving?
lara [203]

Answer:They may cause motion to change

Explanation:They may cause motion; they may also slow, stop, or change the direction of motion of an object that is already moving. Since force cause changes in the speed or direction of an object, we can say that forces cause changes in velocity. Remember that acceleration is a change in velocity.

7 0
3 years 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
A 42.2 kg sled is pulled forward
zaharov [31]

The net force on the sledge  is 31.64N.

Frictional force = µkR

                         = 0.269 x 42.2 x 9.81 = 111.36

net force = 143N - 111.36N

               = 31.64N

refer  brainly.com/question/24557767

#SPJ2

     

7 0
2 years ago
Which part of an object's rate of change best defines acceleration?
grandymaker [24]
I believe it is speed.

Hope this helps!
5 0
3 years ago
Read 2 more answers
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