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insens350 [35]
3 years ago
5

A baseball is hit with a bat. The direction of the ball is completely reversed and its speed is doubled. If the actual contact w

ith the bat lasts what is the ratio of the acceleration to the original velocity?
Physics
2 answers:
alexdok [17]3 years ago
8 0
Before the impact, let the velocity of the baseball was v m/s.

After being hit by the bat its velocity is -2v
So, change in velocity, Deltav=v-(-2v)=3v
Acceleration is defined as the rate of change in velocity, i.e. actual change in velocity divided by the time taken to change it. Time taken to change velocity is the time of actual contact of the bat and ball, i.e. 0.31 s.

a=(Deltav)/(Deltat)
=(3v)/0.37
Therefore, a/v=3/0.31=9.7 s^-1
So, the ratio of acceleration of the baseball to its original velocity is 9.7.
Lubov Fominskaja [6]3 years ago
6 0

Answer:

ratio = 9.67

Explanation:

given,

initial velocity = v

final velocity  = - 2 v

change in velocity = v-(-2v)

                               = 3 v

assuming the time be t = 0.31 s

acceleration = \dfrac{change\ in\ velocity}{time}

                     = \dfrac{3v}{0.31}

                     = 9.67 v

ratio = \dfrac{a}{v} = \dfrac{9.67 v}{v}

ratio = 9.67

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alina1380 [7]

Answer:

1.758820×10^11(-2.5i-0.8j) m/s^2

Explanation:

From the question, the parameters given are; E=(2.80i+ 5.20j) v/m, a uniform magnetic field,B= 0.400K T, acceleration, a= ??? and velocity vector, v= 11.0i metre per seconds (m/s)...

We can solve this problem using the formula below;

Ma= q[E+V × B] ---------------(1).

Note: q is negative, m= mass of electron.

Making acceleration,a the subject of the formula and substituting the parameters into equation (1);

a= -e/m × (2.5i + 5.2j +11.0i × 0.400K)

a= -e/m × (2.5i+5.2j-4.4j)

a= e/m × (-2.5i - 0.8j)

e/m= 1.758820×10^11 c/kg

Therefore, slotting in the value of charge to mass(e/m) ratio;

a= 1.7588×10^11×(-2.5i-0.8j) m/s^2

7 0
3 years ago
Please answer :&gt;<br> 40 POINTS
LenaWriter [7]

Answer:

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8 0
2 years ago
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What is the separation distance, in meters, between masses m1 = 15 x 107 kg and m2 = 62 x 107 kg when the gravitational force be
nalin [4]

Answer:

94.1 m

Explanation:

From Coulombs law,

F  = Gm1m2/r²................... Equation 1

where F = force, m1 = first mass, m2 = second mass, G = universal constant, r = distance of separation.

Make r the subject of the equation,

r = √(Gm1m2/F)................. Equation 2

Given: F = 7×10² N, m1 = 15×10⁷ kg, m2 = 62×10⁷ kg,

Constant: G = 6.67×10⁻¹¹ Nm²/kg²

Substitute into equation 2

r = √( 6.67×10⁻¹¹×15×10⁷×62×10⁷/7×10²)

r √(886.16×10)

r √(88.616×10²)

r = 9.41×10

r = 94.1 m.

Hence the distance of separation = 94.1 m

3 0
3 years ago
Consider a railroad bridge over a highway. A train passing over the bridge dislodges a loose bolt from the bridge, which proceed
photoshop1234 [79]

Answer:

The railroad tracks are 13 m above the windshield (12 m without intermediate rounding).

Explanation:

First, let´s calculate the time it took the driver to travel the 27 m to the point of impact.

The equation for the position of the car is:

x = v · t

Where

x = position at time t

v = velocity

t = time

x = v · t

27 m = 17 m/s · t

27 m / 17 m/s = t

t = 1.6 s

Now let´s calculate the distance traveled by the bolt in that time. Let´s place the origin of the frame of reference at the height of the windshield:

The position of the bolt will be:

y = y0 + 1/2 · g · t²

Where

y = height of the bolt at time t

y0 = initial height of the bolt

g = acceleration due to gravity

t = time

Since the origin of the frame of reference is located at the windshield, at time 1.6 s the height of the bolt will be 0 m (impact on the windshield). Then, we can calculate the initial height of the bolt which is the height of the railroad tracks above the windshield:

y = y0 + 1/2 · g · t²

0 = y0 -1/2 · 9.8 m/s² · (1.6 s)²

y0 = 13 m

8 0
3 years ago
The rubidium isotope 87Rb is a β emitter that has a half-life of 4.9 ✕ 1010 y that decays into 87Sr. It is used to determine the
fredd [130]

Answer:

t = 39.04 1010 year

Explanation:

This is a nuclear disintegration exercise that is governed by the equation.

     N = N0 e (-lam t)

The average life time is related to nuclear activity

      T ½ = ln 2 / lam

Let's use these two equations for exercise, let's start by finding nuclear activity

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      No / N = 0.0040

Let's look

       No / N = 1/0040

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Let's calculate the time

       (-lam t) = ln (N / No)

,        t = - 1 / lam ln (n / No)

        t = - 1 / 0.14146 10-10 ln (250)

        t = 39.04 1010 year

4 0
2 years ago
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