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zubka84 [21]
3 years ago
6

You're driving down the highway late one night at 16.0 m/s when a deer steps onto the road 39.0 m in front of you. your reaction

time before stepping on the breaks is 0.50 s, and the maximum deceleration of your car is 12.0 m/s2 . how much distance is in between you and the deer when you come to a stop? what is the maximum speed you could have and still not hit the dear?
Physics
1 answer:
choli [55]3 years ago
5 0
The breaking distance consists of two parts. The first part is the first 0.5 seconds were no breaking occurs. Given values: t time, v₀ initial velocity:
x₁ = v₀*t
The second part occurs after t = 0,5s with the given acceleration: a = - 12 m/s²
were the final velocity is zero, v = 0 and the initial velocity v₀= 16m/s:
v = a*t + v₀ = 0 => v₀ = -a*t => t = v₀/-a

x₂ = 0.5*a*t² = 0.5*v°²/a

The total breaking distance is the sum of the two parts:
x = x₁ + x₂ = v₀* t + 0.5 * v₀² / a = 16 * 0.5 + 0.5 * 16² / 12 = 8 + 10,7 = 18,7

You can use this result to calculate the remaining distance. You can use the last equation to calculate the maximum speed you could have to avoid a collision.
Use x = 39m and solve for v₀. 
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Over a span of 6.0 seconds, a car changes it's speed from 89 km/h to 37 km/h. What is its average acceleration in meters per sec
scoundrel [369]

Acceleration = (change in speed) / (time for the change)

change in speed = (speed at the end) - (speed at the beginning)

change in speed = (37 km/hr) - (89 km/hr) = -52 km/hr

Acceleration = (-52 km/hr) / (6 sec)

Acceleration = (-26/3) km/(hr·sec)

Units: (1/hr·sec) · (hr/3600 sec) = 1 / 3600 sec²

(-26/3) km/(hr·sec) = (-26/3) km/(3600 sec²)

= -26,000/(3 · 3600) m/s²

<em>Acceleration = -2.41 m/s²</em>

3 0
3 years ago
You throw a ball vertically upward, and as it leaves your hand, its speed is 10.0 m/s.
8090 [49]
The answer is A bc I did the quiz and I got it right
5 0
3 years ago
Cheetahs can accelerate to a speed of 19.6 m/s in 2.45 s and can continue to accelerate to reach a top speed of 27.6 m/s . Assum
marta [7]

Answer:

  • v_{top} = 61.96 \frac{mi}{h}

Explanation:

To express the cheetah's top speed in miles per hour, we just need to find the conversion factor.

We know that the top speed is

v_{top} = 27.7 \frac{m}{s}

So, we want to obtain miles from meters and hours from seconds.

<h3>miles from meters</h3>

First we write the equivalence:

1609.34 \ m = 1 \ mi

Now, we can divide by 1609.34 meters on both sides:

\frac{1609.34 \ m}{ 1609.34 \ m} = \frac{1 \ mi}{ 1609.34 \ m}

The left sides equals 1, so

1 = \frac{1 \ mi}{ 1609.34 \ m}

And this is our conversion factor from meters to miles. Now, we can multiply our top speed by this conversion factor, as the conversion factor equals one, and is dimensionless, the physical meaning will be the same.

v_{top} = 27.7 \frac{m}{s} * \frac{1 \ mi}{ 1609.34 \ m}

v_{top} = 27.7 \frac{m}{s} * \frac{1 \ mi}{ 1609.34 \ m}

v_{top} = 0.0172120 \frac{mi}{s}

This is the top speed in miles per second, now, for obtaining miles per hour:

<h3>hours from seconds</h3>

We can do pretty much the same, first, the equivalence:

1 \ h = 3600 \ s

as the seconds are dividing in the velocity, we know divide by 1 hour.

\frac{1 \ h}{ 1 \ h} = \frac{3600 \ s}{ 1 \ h}

1 = \frac{3600 \ s}{ 1 \ h}

and know we just multiply our top speed by this conversion factor

v_{top} = 0.0172120 \frac{mi}{s}  \frac{3600 \ s}{ 1 \ h}

v_{top} = 61.96 \frac{mi}{h}

8 0
3 years ago
a body initially at rest, starts moving with a constant acceleration of 2ms-2 .calculate the velocity acquired and the distance
Marta_Voda [28]

a) 10 m/s

b) 25 m

Explanation:

a)

The body is moving with a constant acceleration, therefore we can solve the problem by using the following suvat equation:

v=u+at

where

u is the initial velocity

v is the final velocity

a is the acceleration

t is the time

For the body in this problem:

u = 0 (the body starts from rest)

a=2 m/s^2 is the acceleration

t = 5 s is the time

So, the final velocity is

v=0+(2)(5)=10 m/s

b)

In this second part, we want to calculate the distance travelled by the body.

We can do it by using another suvat equation:

v^2-u^2=2as

where

u is the initial velocity

v is the final velocity

a is the acceleration

s is the distance travelled

Here we have

u = 0 (the body starts from rest)

a=2 m/s^2 is the acceleration

v = 10 m/s is the final velocity

Solving for s,

s=\frac{10^2-0^2}{2(2)}=25 m

3 0
3 years ago
Nuclearissionoccursthroughmanydifferent pathways. For the ission of U-235 induced by a neutron, write a nuclear equation to form
ruslelena [56]

Answer: a) ^{235}_{92}\textrm{U}+^{1}_{0}\textrm{n} \rightarrow ^{87}_{35}\textrm {Br}+ ^{146}_{57}\La + 3^{1}_{0}n

b) ^{235}_{92}\textrm{U}+^1_0\textrm{n}\rightarrow ^{140}_{56}\textrm{Ba}+^{94}_{36}\textrm{Kr}+2^1_0\textrm{n}

Explanation:

A nuclear fission reaction is defined as the reaction in which a heavy nucleus splits into small nuclei along with release of energy.

a) The given reaction is ^{235}_{92}\textrm{U}+^{1}_{0}\textrm{n} \rightarrow ^{87}_{35}\textrm {Br}+ ^{146}_{57}\La + x^{1}_{0}n

Now,  as the mass on both reactant and product side must be equal:

235+1=87+146+x

x=3

Thus three neutrons are produced and nuclear equation will be: ^{235}_{92}\textrm{U}+^{1}_{0}\textrm{n} \rightarrow ^{87}_{35}\textrm {Br}+ ^{146}_{57}\La + 3^{1}_{0}n

b) For the another fission reaction:

^{235}_{92}\textrm{U}+^1_0\textrm{n}\rightarrow ^{A}_{56}\textrm{Ba}+^{94}_{Z}\textrm{X}+2^1_0\textrm{n}

To calculate A:

Total mass on reactant side = total mass on product side

235 + 1 = A + 94 + 2

A = 140

To calculate Z:

Total atomic number on reactant side = total atomic number on product side

92 + 0 = 56 + Z + 0

Z = 36

As Krypton has atomic number of 36,Thus the nuclear equation will be :

^{235}_{92}\textrm{U}+^1_0\textrm{n}\rightarrow ^{140}_{56}\textrm{Ba}+^{94}_{36}\textrm{Kr}+2^1_0\textrm{n}

         

8 0
3 years ago
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