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kirza4 [7]
3 years ago
11

A car of mass 1470 kg is on an icy driveway

Physics
1 answer:
Schach [20]3 years ago
8 0

Answer:

  a = 7.5 m / s²

Explanation:

For this exercise let's use Newton's second law, let's create a coordinate system with the x axis parallel to the plane and the y axis perpendicular to the plane

Y axis

       N - W cos θ = 0

       N = mg cos θ

X axis

       W sin θ = m a

 

       mg sin θ = m a

        a = g sin θ

let's calculate

        a = 9.8 cos 40

        a = 7.5 m / s²

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A 42.2 kg sled is pulled forward
zaharov [31]

The net force on the sledge  is 31.64N.

Frictional force = µkR

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net force = 143N - 111.36N

               = 31.64N

refer  brainly.com/question/24557767

#SPJ2

     

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2 years ago
Magnesium hydroxide is used in? <br><br> soap<br><br> batteries<br><br> antacids<br><br> detergents
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3 years ago
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A wave travels with speed 200 m/s. Its wave number is 1.5 rad/m. What are its (a) wavelength and (b) frequency
Vladimir [108]

Answer:

(a)

Explanation:

5 0
3 years ago
A sailboat weighing 980 lb with its occupants is running downwind at 8 mi/h when its spinnaker is raised to increase its speed.
Effectus [21]

Answer:

78.498N

Explanation:

The Net force provided by the spinnaker can be obtained from Newton's second law of motion as follows;

F=\frac{m(v-u)}{t}................(1)

where m is the mass, v is the final velocity, u is the initial velocity and t is the time interval for which the force acted.

Given;

m =980lb

v = 12mi/h

u =8mi/hr

t = 10s.

It is important to convert all quantities to their SI units where necessary, so we do that as follows;

1lb = 0.45kg,

hence 980lb = 980 x 0.45kg = 441kg.

1mile = 1609.34m

1hour = 3600s,

therefore;

8mi/h=\frac{8*1609.34m}{3600s}=3.58m/s

12mi/h=\frac{12*1609.34m}{3600s}=5.36m/s

Substituting all values into equation (1), we obtain the following;

F=\frac{441(5.36-3.58)}{10}\\F=\frac{441*1.78}{10}\\F=\frac{784.98}{10}\\F=78.498N

4 0
3 years ago
A 5.00 kg crate is suspended from the end of a short vertical rope of negligible mass. An upward force F(t) is applied to the en
Brut [27]

Answer:

75 N

Explanation:

In this problem, the position of the crate at time t is given by

y(t)=2.80t+0.61t^3

The velocity of the crate vs time is given by the derivative of the position, so it is:

v(t)=y'(t)=\frac{d}{dt}(2.80t+0.61t^3)=2.80+1.83t^2

Similarly, the acceleration of the crate vs time is given by the derivative of the velocity, so it is:

a(t)=v'(t)=\frac{d}{dt}(2.80+1.83t^2)=3.66t [m/s^2]

According to Newton's second law of motion, the force acting on the crate is equal to the product between mass and acceleration, so:

F(t)=ma(t)

where

m = 5.00 kg is the mass of the crate

At t = 4.10 s, the acceleration of the crate is

a(4.10)=3.66\cdot 4.10 =15.0 m/s^2

And therefore, the force on the crate is:

F=ma=(5.00)(15.0)=75 N

7 0
3 years ago
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