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Ann [662]
3 years ago
6

With what force will a car hit a tree if the car has a mass of 3,550 kg and it is accelerating at a rate of 2.5 m/s2 on a snowy

day when the temperature drops to 31oC?
Physics
1 answer:
bekas [8.4K]3 years ago
3 0

Answer:

<h2>8875 N</h2>

Explanation:

The force acting on an object given it's mass and acceleration can be found by using the formula

force = mass × acceleration

From the question we have

force = 3550 × 2.5

We have the final answer as

<h3>8875 N</h3>

Hope this helps you

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An 80 g, 40 cm long rod hangs vertically on a frictionless, horizontal axle passing through its center. A 15 g ball of clay trav
fiasKO [112]

Answer:

θ  = 12.95º

Explanation:

For this exercise it is best to separate the process into two parts, one where they collide and another where the system moves altar the maximum height

Let's start by finding the speed of the bar plus clay ball system, using amount of momentum

The mass of the bar (M = 0.080 kg) and the mass of the clay ball (m = 0.015 kg) with speed (v₀ = 2.0 m / s)

Initial before the crash

      p₀ = m v₀

Final after the crash before starting the movement

     p_{f} = (m + M) v

     p₀ = p_{f}

     m v₀ = (m + M) v

     v = v₀ m / (m + M)

     v = 2.0 0.015 / (0.015 +0.080)

     v = 0.316 m / s

With this speed the clay plus bar system comes out, let's use the concept of conservation of mechanical energy

Lower

    Em₀ = K = ½ (m + M) v²

Higher

    E_{mf} = U = (m + M) g y

   Em₀ = E_{mf}

   ½ (m + M) v² = (m + M) g y

   y = ½ v² / g

   y = ½ 0.316² / 9.8

   y = 0.00509 m

Let's look for the angle the height from the pivot point is

    L = 0.40 / 2 = 0.20 cm

The distance that went up is

     y = L - L cos θ

     cos θ  = (L-y) / L

     θ  = cos⁻¹ (L-y) / L

     θ  = cos⁻¹-1 ((0.20 - 0.00509) /0.20)

      θ  = 12.95º

4 0
3 years ago
I need to find 1).a,b,c
Aleksandr [31]
Let's cut through the weeds and the trash
and get down to the real situation:

                  A stone is tossed straight up at  5.89 m/s .
                  Ignore air resistance.


Gravity slows down the speed of any rising object by  9.8 m/s every second.
So the stone (aka Billy-Bob-Joe) continues to rise for

                     (5.89 m/s / 9.8 m/s²)  =  0.6 seconds.

At that timer, he has run out of upward gas.  He is at the top
of his rise, he stops rising, and begins to fall.

His average speed on the way up is  (1/2) (5.89 + 0) = 2.945 m/s .

Moving for 0.6 seconds at an average speed of  2.945 m/s,
he topped out at

                    (2.945 m/s) (0.6 s) =  1.767 meters above the trampoline.

With no other forces other than gravity acting on him, it takes him
the same time to come down from the peak as it took to rise to it.

   (0.6 sec up) + (0.6 sec down)  =  1.2 seconds until he hits rubber again.



 
5 0
4 years ago
How fast does a 500 Hz wave travel if its wavelength is 0.5 m?
olga_2 [115]

Answer:

250 m/s

Explanation:

4 0
3 years ago
Read 2 more answers
1.Mention two uses of the concave mirror.
Anuta_ua [19.1K]

Answer:

1. telescope

2.

f =  \frac{r}{2}

f =  \frac{r}{2}f- focal length

f =  \frac{r}{2}f- focal length r- the radius of curvature of the mirror

\frac{1}{f}  =   \frac{1}{p}  +  \frac{1}{l}

p-the distance of the object from the vertex of the mirror

l-the distance of the figure from the vertex of the mirror

8 0
3 years ago
A ship sets sail from Rotterdam, The Netherlands, intending to head due north at 6.5 m/s relative to the water. However, the loc
sattari [20]

Answer:

Explanation:

velocity of ship with respect to water = 6.5 m/s due north

\overrightarrow{v}_{s,w}=6.5 \widehat{j}

velocity of water with respect to earth = 1.5 m/s at 40° north of east

\overrightarrow{v}_{w,e}=1.5\left ( Cos40\widehat{i} +Sin40\widehat{j}\right)

velocity of ship with respect to water = velocity of ship with respect to earth - velocity of water with respect to earth

\overrightarrow{v}_{s,w} = \overrightarrow{v}_{s,e} - \overrightarrow{v}_{w,e}

\overrightarrow{v}_{s,e} = 6.5 \widehat{j}- 1.5\left (Cos40\widehat{i} +Sin40\widehat{j}  \right )

\overrightarrow{v}_{s,e} = - 1.15 \widehat{i}+5.54\widehat{j}

The magnitude of the velocity of ship relative to earth is \sqrt{1.15^{2}+5.54^{2}} = 5.66 m/s

5 0
3 years ago
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