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3 years ago

A car slid off an icy 10m bridge and landed 12m away from the bridge. How much time was the car in the air? (Hunt: Projectile)

Physics

2 answers:

Contact [7]3 years ago

8 0

You will use the height of the bridge from the ground.

Solution:

Formula to be used is y=Viy(t)+g(t^2)/2

Where:

Vi=initial velocity which is 0 m/s

y=10 m

Gravitational acceleration or g =9.8m/s^2

T= time you need

Substitute all the given to the formula

10m=(0m/s)(t)+(9.8m/s^2)(t^2)/2

10mx2=9.8m/s^2(t^2)

Now isolate the variable you want to find which is T or time

10mx2/9.8m/s^2=t^2

20m/9.8m/s^2=t^2

Square root of 2.04= square root of t^2

T=1.43 secs

The answer is 1.43 seconds

STatiana [176]3 years ago

3 0

You just need to use the height of the bridge from the ground.

The formula to be used is y = Voy*t + g *(t^2) / 2, where the Voy is the intitial vertical velocity, which is zero.

Then, y = 10 m = g * (t^2) / 2 => t^2 = 2 * 10 m / g

=> t^2 = 2 * 10 m / 9.8 m/s^2 = 2.04 s^2

=> t = 1.43 s <------- answer

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You put a bottle of soft drink in a refrigerator and leave it until its temperature has dropped 12.0K .

Bingel [31]

Answer: a) - 437.8° F, b) - 261°c.

Explanation: a) the kelvin and Fahrenheit temperature scale are related by the formulae below.

5 (°F - 32) = 9 (k - 273)

Where °F = temperature in Fahrenheit and k = temperature in kelvin.

For question A, k = 12.0, by substituting to have the value for °F, we have

5(°F - 32) = 9 ( 12 - 273)

5(°F - 32) = 9(-261)

5(°F - 32) = - 2349

°F - 32 = - 2349/5

°F - 32 = - 469.8

°F = - 469.8 + 32

°F = - 437.8

Question B

The centigrade and kelvin scale are related by the formulae below

°c = k - 273

Where °c = temperature in centigrade and k = temperature in kelvin =12

°c = 12 - 273

°c = - 261

3 0

3 years ago

A point charge q1 = -4.00 nC is at the point x = 0.60 m, y = 0.80 m , and a second point charge q2 = +6.00 nC is at the point x

makkiz [27]

Answer:

A)Magnitude of the net electric field at the origin due q₁ and q₂

E₀= 131.6 N/C

B) Direction of the net electric field at the origin due q₁ and q₂

β=3.91°

Explanation:

Conceptual analysis

The electric field at a point P due to a point charge is calculated as follows:

E = k*q/d²

E: Electric field in N/C

q: charge in Newtons (N)

k: electric constant in N*m²/C²

d: distance from charge q to point P in meters (m)

The electric field at a point P due to several point charges is the vector sum of the electric field due to individual charges.

Equivalences

1nC= 10⁻⁹ C

Data

q₁ = -4 nC = -4*10⁻⁹ C

q₂ = +6nC =+ 6*10⁻⁹ C

k = 9*10⁹ N*m²/C²

$d_{1} =\sqrt{0.6^{2}+0.8^{2} } =1 m$

d₂ = 0.6 m

Graphic attached

The attached graph shows the field in x=0, y=0, due to the charges q₁ and q₂:

E₁: Total field at point x=0 , y=0 due to charge q₁. As the charge q₁ is negative (q₁-), the field enters the charge.

E₂: Total field at point x=0 , y=0 due to charge q₂. As the charge q2 is positive (q₂+) ,the field leaves the charge.

Calculation of the electric field at the origin of x-y coordinates due to the charge q₁

E₁=K*q₁/d₁²=9*10⁹*4*10⁻⁹/1²=36N/C

Components (x-y) of the field due to q1:

E₁x=E₁*cosα = 36*(0.6/1) = 36*(0.6) = 21.6 N/C

E₁y=E₁*sinα = 36*(0.8/1) = 36*(0.8) =28.8 N/C

Calculation of the electric field at the origin of x-y coordinates due to the charge q₂

E₂=K*q₂/d₂² = -9*10⁹*6*10⁻⁹/0.6² = -150 N/C

Calculation of the electric field components at the origin of x-y coordinates

E₀ is the electric field at the origin of x-y coordinates (x=0,y=0)

E₀x= E₁x+E₂= 21.6-150 = -128.4 N/C

E₀y= E₁y = 28.8 N/C

A )Magnitude of the net electric field at the origin due q₁ and q₂

$E_{o} =\sqrt{128.4^{2}+28.8^{2} }$

E₀= 131.6 N/C

B) Direction of the net electric field at the origin due q₁ and q₂

$\beta =tanx{-1} \frac{E_{oy} }{Eox}$

$\beta =tan^{-1} \frac{28.8}{128.4}$

β=3.91°

6 0

3 years ago

a bowling ball is pushed with a force of 22.0 n and acclerates at 5.5m/s2. what mass of of the bwling ball

inn [45]

A force can cause a resting object to move, or it can accelerate a moving object by changing the objects speed or direction

4 0

3 years ago

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Two spherical asteroids have the same radius R. Asteroid 1 hasmass M and asteroid 2 has mas 2M. The two asteroids are releasedfr

gtnhenbr [62]

Answer:

$v_1 =\sqrt{\dfrac{16GM}{15R}}$

$v_2 =\sqrt{\dfrac{4GM}{15R}}$

Explanation:

given,

mass of asteroid 1 = M

mass of asteroid 2 = 2M

radius of two asteroid = R

Distance between the asteroid = 10 R

Speed of the asteroid before collision = ?

using conservation of momentum

M u + 2M u' = M v₁ + 2 M v₂

initial speed of asteroid is equal to zero

0 = v₁ + 2 v₂

v₁ = -2 v₂

using conservation of momentum

initial potential energy is converted into potential energy and the kinetic energy of both the asteroids.

$\dfrac{GM(2M)}{10R}=\dfrac{GM(2M)}{2R}+\dfrac{1}{2}Mv_1^2 + \dfrac{1}{2}(2M)v_2^2$

$\dfrac{GM(2M)}{10R}-\dfrac{GM(2M)}{2R}=\dfrac{1}{2}M(-2v_2)^2 + \dfrac{1}{2}(2M)v_2^2$

$6v_2^2 = \dfrac{8GM}{5R}$

$v_2 =\sqrt{\dfrac{4GM}{15R}}$

now,

$v_1 =-2\sqrt{\dfrac{4GM}{15R}}$

$v_1 =\sqrt{\dfrac{16GM}{15R}}$

hence, the velocity of asteroid are

$v_1 =\sqrt{\dfrac{16GM}{15R}}$

$v_2 =\sqrt{\dfrac{4GM}{15R}}$

6 0

3 years ago

Suppose you walk 17 m straight west and then 26.5 m straight north. What is the compass direction, in degrees measured West of N

garik1379 [7]

Answer:32.7

Explanation:

You must first make a right triangle with the two legs being 17 m west and 26.5 m north. To find the degree measurement west of north you must find the angle between the north leg and the hypotenuse. then just use inverse tangent of (17/26.5) which is 32 .7 degrees

6 0

3 years ago