Joe places two solid objects in contact with each other.

Two cars A and B are 100m apart moving towards each other with

maxonik [38]

Let car A's starting position be the origin, so that its position at time t is

A: x = (40 m/s) t

and car B has position at time t of

B: x = 100 m - (60 m/s) t

They meet when their positions are equal:

(40 m/s) t = 100 m - (60 m/s) t

(100 m/s) t = 100 m

t = (100 m) / (100 m/s) = 1 s

so the cars meet 1 second after they start moving.

They are 100 m apart when the difference in their positions is equal to 100 m:

(40 m/s) t - (100 m - (60 m/s) t) = 100 m

(subtract car B's position from car A's position because we take car A's direction to be positive)

(100 m/s) t = 200 m

t = (200 m) / (100 m/s) = 2 s

so the cars are 100 m apart after 2 seconds.

3 0

3 years ago

A 7.0kg object rests on a horizontal frictionless surface. What is the magnitude of the horizontal

sukhopar [10]

Answer:

16.1 N

Explanation:

From the question,

F = ma.............................. Equation 1

Where F = horizontal force, m = mass of the object, a = acceleration .

Given: m = 7.0 kg, a = 2.3 m/s²

Substitute this values into equation 1

F = (7.0×2.3)

F = 16.1 N.

Hence the magnitude of the horizontal force is 16.1 N

6 0

3 years ago

At t = 0, a particle starts at rest and moves along a line in such a way that, at time t, its acceleration is 24t 2 ft / s2. thr

svp [43]

The acceleration of the particle at time t is:
$a(t)=24t^2 ft/s^2$
The velocity of the particle at time t is given by the integral of the acceleration a(t):
$v(t)= \int a(t) \, dt = \int (24 t^2) dt=24 \frac{t^3}{3}=8t^3 ft/s$
and the position of the particle at time t is given by the integral of the velocity v(t):
$x(t)=\int v(t) = \int (8t^3)=8 \frac{t^4}{4}=2t^4 ft$

Assuming the particle starts from position x(0)=0 at t=0, the distance the particle covers in the first t=2 seconds can be found by substituting t=2 s in the equation of x(t):
$x(2 s)=2 t^4 = 2 (2s)^4=32 ft$

5 0

3 years ago

Please help need answer asp

fenix001 [56]

The Bio-Mechanical term that defines managing your force while maintaining balance is "Stability"

8 0

4 years ago

Read 2 more answers

A very long straight wire has charge per unit length 1.44×10-10C/m.

4vir4ik [10]

Answer:

Distance of the point where electric filed is 2.45 N/C is 1.06 m

Explanation:

We have given charge per unit length, that is liner charge density $\lambda =1.44\times 10^{-10}C/m$

Electric field E = 2.45 N/C

We have to find the distance at which electric field is 2.45 N/C

We know that electric field due to linear charge is equal to

$E=\frac{\lambda }{2\pi \epsilon _0r}$ , here $\lambda$ is linear charge density and r is distance of the point where we have to find the electric field

So $2.45=\frac{1.44\times 10^{-10} }{2\times 3.14\times 8.85\times 10^{-12}\times r}$

r = 1.06 m

So distance of the point where electric filed is 2.45 N/C is 1.06 m

3 0

3 years ago