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zaharov [31]
2 years ago
13

2. How long would it take a car to reach 50 m/s if they accelerated at a constant rate of 5.44

Physics
1 answer:
Mariulka [41]2 years ago
6 0

9.19 seconds (2dp)

Given:

v = 50m/s

u = 0 m/s

a = 5.44 m/s

t = ?

Substitute values into the equation v = u + at

v = u + at

50 = 0 + 5.44 x t

t = 50/5.44

t = 9.19 seconds

Time is in seconds as the velocity and acceleration are in metres per second instead of kilometres per hour

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An apple hanging from a limb has potential energy because of its height . If it falls ,what becomes of this energy just before i
sineoko [7]

Answer: Just before its hits the ground it becomes kinetic energy and when it hits the ground it becomes in another form of energy (acoustic energy or thermal energy, for example)

Explanation:

Energy is the ability of matter to produce work in the form of movement, light, heat, among others.

In this sense, according to the Conservation of Energy principle: <em>"energy is not created or destroyed, it is transformed."</em>

So, in the case of the apple, its total energy is conserved.

When the apple is hanging from a limb, it has zero kinetic energy K_{o}=0 (because it is at rest) and has gravitational potential energy U_{o}, which depends on the mass m, the acceleration due gravity g and the height h:

U_{o}=mgh

When the apple falls, just before its hits the ground, this gravitational potential energy transforms in kinetic energy K_{1} (since the apple is moving), which depends on the mass and velocity V of the apple:

K_{1}=\frac{1}{2}mV^{2}

When the apple hits the ground, the gravitational potential energy is zero (h=0) and the kinetic energy transforms into some other form of energy (acoustic energy or thermal energy, for example).

3 0
3 years ago
A 45.0 kg skier testing a new competitive ski wax drops off a ledge, goes into a crouch and goes straight down a slope of 10.0◦
Neko [114]

Answer:

v = 21.03 m/s

Explanation:

given,

mass of skier = 45 kg        

the slope of the snow = 10.0◦      

coefficient of friction = 0.114  

distance traveled = 300 m      

speed = ?              

Acceleration = g sin θ - µ g Cos θ        

= 9.8 × Sin (10°) - 0.10 × 9.8 × Cos(10°)        

= 0.737 m/s²      

using equation of motion        

v² = u² + 2 a s        

v² = 0 + 2 × 0.737 × 300            

v = 21.03 m/s                  

Speed of skier's after travelling 300 m speed is  equal to 21.03 m/s

6 0
3 years ago
A particle leaves the origin with a speed of 3.6 106 m/s at 34 degrees to the positive x axis. It moves in a uniform electric fi
Andrej [43]

Answer:

E = -4556.18 N/m

Explanation:

Given data

u = 3.6×10^6 m/sec

angle = 34°

distance x = 1.5 cm = 1.5×10^-2 m  (This data has been assumed not given in

Question)

from the projectile motion the horizontal distance traveled by electron is

x = u×cosA×t

⇒t = x/(u×cos A)

We also know that force in an electric field is given as

F = qE

q= charge , E= strength of electric field

By newton 2nd law of motion

ma = qE

⇒a = qE/m

Also, y = u×sinA×t - 0.5×a×t^2

⇒y = u×sinA×t - 0.5×(qE/m)×t^2

if y = 0 then

⇒t = 2mu×sinA/(qE) = x/(u×cosA)

Also, E = 2mu^2×sinA×cosA/(x×q)

Now plugging the values we get

E = 2×9.1×10^{-31}×3.6^2×10^{12}×(sin34°)×(cos34°)/(1.5×10^{-2}×(-1.6)×10^{-19})

E = -4556.18 N/m

4 0
3 years ago
Light of wavelength 600 nm passes though two slits separated by 0.22mm and is observed on a screen 1.1m behind the slits. The lo
Ne4ueva [31]

Answer:

y(m=1)=\frac{(1)(600*10^{-9}m)(1.1m)}{0.22m}=3*10^{-6}m\\\\y(m=1)=\frac{(-1)(600*10^{-9}m)(1.1m)}{0.22m}=-3*10^{-6}m

Explanation:

We have to take into account the expression for the position of the fringes

dsen\theta=m\lambda\\y=\frac{m\lambda D}{d}

where m is the number of the maximum, d is the separation of the slits, D is the distance to the screen.

(a) By replacing we obtain

y(m=1)=\frac{(1)(600*10^{-9}m)(1.1m)}{0.22m}=3*10^{-6}m\\\\y(m=1)=\frac{(-1)(600*10^{-9}m)(1.1m)}{0.22m}=-3*10^{-6}m

(b)  more information is required to solve this point. Please complete the information.

HOPE THIS HELPS!

4 0
3 years ago
Read 2 more answers
I appreciate the help on this physics question
Sidana [21]
Not enough information
3 0
2 years ago
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