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2 years ago

If an object is accelerating, it is traveling the same distance for each time interval of its motion

Physics

2 answers:

Elenna [48]2 years ago

6 0

The answer is false :)

Anika [276]2 years ago

4 0

Answer:

false

Explanation:

if an object is accelerating the object will not travel the same distance every time interval

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What question was asked by Faraday that the narrator calls a leap

masha68 [24]

<span> If electricity and magnetism can create motion, can the reverse be true? Can motion and magnetism create electricity?</span>

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3 years ago

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Which state of matter has a definite volume but no definite shape

a_sh-v [17]

The answer is liquid.

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3 years ago

If a force is 100 N and is pointing 37 degrees north of east. (a) Draw a diagram of this force. (b) Draw the force's x and y com

sasho [114]

Answer

given,

force = 100 N

Point 37 degrees north of east

a) and b) part is shown in the diagram attached below.

c) to find the x and y component of the force

x- component of the force

$F_x = F cos \theta$

$F_x = 100\times cos 37^0$

$F_x = 79.86 N$

y- component of the force

$F_y = F sin \theta$

$F_y = 100\times sin 37^0$

$F_y = 60.18 N$

3 0

3 years ago

2. An athlete of average size is hanging from the end of a 20 m long rope, which has a mass of 4 kg and is attached to a hook in

a_sh-v [17]

Answer:

t = 0.319 s

Explanation:

With the sudden movement of the athlete a pulse is formed that takes time to move along the rope, the speed of the rope is given by

v = √T/λ

Linear density is

λ = m / L

λ = 4/20

λ = 0.2 kg / m

The tension in the rope is equal to the athlete's weight, suppose it has a mass of m = 80 kg

T = W = mg

T = 80 9.8

T = 784 N

The pulse rate is

v = √(784 / 0.2)

v = 62.6 m / s

The time it takes to reach the hook can be searched with kinematics

v = x / t

t = x / v

t = 20 / 62.6

t = 0.319 s

7 0

3 years ago

A 1400 kg car starts from rest on a horizontal road and gains a speed of 61 km/h in 19 s. (a) what is its kinetic energy at the

lana [24]

(a) Let's convert the final speed of the car in m/s:
$v_f = 61 km/h = 16.9 m/s$
The kinetic energy of the car at t=19 s is
$K= \frac{1}{2}mv_f^2= \frac{1}{2}(1400 kg)(16.9 m/s)^2=2.00 \cdot 10^5 J$

(b) The average power delivered by the engine of the car during the 19 s is equal to the work done by the engine divided by the time interval:
$P= \frac{W}{\Delta t}$
But the work done is equal to the increase in kinetic energy of the car, and since its initial kinetic energy is zero (because the car starts from rest), this translates into
$P= \frac{K}{\Delta t}= \frac{2.00 \cdot 10^5 J}{19 s}=1.05 \cdot 10^4 W$

(c) The instantaneous power is given by
$P_i = Fv_f$
where F is the force exerted by the engine, equal to F=ma.

So we need to find the acceleration first:
$a= \frac{v_f-v_i}{\Delta t}= \frac{16.9 m/s}{19 s}=0.89 m/s^2$
And the problem says this acceleration is constant during the motion, so now we can calculate the instantaneous power at t=19 s:
$P_i = Fv=(ma)v=(1400 kg)(0.89 m/s^2)(16.9 m/s)=2.11 \cdot 10^4 W$

5 0

3 years ago