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Nesterboy [21]
3 years ago
8

How high can a body vertically thrown with a speed of 40m/s raise after 3 sec (neglecting air

Physics
1 answer:
Tcecarenko [31]3 years ago
5 0

y = 75.9 m

Explanation:

y = -(1/2)gt^2 + v0yt + y0

If we put the origin of our coordinate system at the point where a body is launched, then y0 = 0.

y = -(1/2)(9.8 m/s^2)(3 s)^2 + (40 m/s)(3 s)

= -44.1 m + 120 m

= 75.9

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According to Newton’s Second Law of Motion, an object will accelerate if which kind of force is applied?
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unbalanced force

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Racing greyhounds are capable of rounding corners at very high speeds. A typical greyhound track has turns that are 45-m-diamete
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Explanation:

It is given that,

Diameter of the semicircle, d = 45 m

Radius of the semicircle, r = 22.5 m      

Speed of greyhound, v = 15 m/s

The greyhound is moving under the action of centripetal acceleration. Its formula is given by :

a=\dfrac{v^2}{r}

a=\dfrac{(15)^2}{22.5}

a=10\ m/s^2

We know that, g=9.8\ m/s^2

a=\dfrac{10\times g}{9.8}

a=1.02\ g

Hence, this is the required solution.                                              

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3 years ago
50 POINTS!
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<h3>No:1</h3>

The object is moving with constant or uniform acceleration and in average speed

<h3>No:-2</h3>

The object is de accelerating

<h3>No:-3</h3>

The object deaccelerated and came to rest so fast.

<h3>No:-4</h3>

The object moves slowly first then accelerated.

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7 0
3 years ago
A proton initially moves left to right along the x-axis at a speed of 2.00 x 103 m/s. It moves into an uniform electric field, w
FinnZ [79.3K]

Answer:

E = 1.04*10⁻¹ N/C

Explanation:

Assuming no other forces acting on the proton than the electric field, as this is uniform, we can calculate the acceleration of the proton, with the following kinematic equation:

vf^{2} -vo^{2} = 2*a*x

As the proton is coming at rest after travelling 0.200 m to the right,  vf = 0, and x = 0.200 m.

Replacing this values in the equation above, we can solve for a, as follows:

a = \frac{vo^{2}*mp}{2*x} = \frac{(2.00e3m/s)^{2}}{2*0.2m} = 1e7 m/s2

According to Newton´s 2nd Law, and applying the definition of an electric field, we can say the following:

F = mp*a = q*E

For a proton, we have the following values:

mp = 1.67*10⁻²⁷ kg

q = e = 1.6*10⁻¹⁹ C

So, we can solve for E (in magnitude) , as follows:

E = \frac{mp*a}{e} =\frac{1.67e-27kg*1e7m/s2}{1.6e-19C} = 1.04e-1 N/C

⇒ E = 1.04*10⁻¹ N/C

5 0
3 years ago
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