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uranmaximum [27]
3 years ago
10

Why can't you trust the law of gravity? (RIDDLE)

Physics
2 answers:
alukav5142 [94]3 years ago
8 0
Because it will always let you down
GarryVolchara [31]3 years ago
5 0
It will always let u down I've heard this many times haha
You might be interested in
A certain unbalanced force gives a 5kg object an acceleration of 15 m/s2. What would be the acceleration if the same force was a
stira [4]

Answer:

a2 = 2.5 m/s2

Explanation:

F1 = m1 a1 We use the same force so F1 = F2

= 5kg × 15m/s2 F2 = m2 a2

= 75N a2 is required

a2 = F2 / m2

= 75N / 30 kg

= 2.5 m/s2

7 0
4 years ago
During a 0.001 s interval while it is between the plates, the change of the momentum of the electron Δp with arrow is < 0, -8
Delvig [45]

Answer:

5.5 x 10^5 N/C

Explanation:

t = 0.001 s

Δp = - 8.8 x 10^-17 kg m /s

Force is equal to the rate of change of momentum.

F = Δp / Δt

F = (8.8 x 10^-17) / 0.001 = 8.8 x 10^-14 N

q = 1.6 x 10^-19 C

Electric field, E = F / q = (8.8 x 10^-14) / (1.6 x 10^-19)

E = 5.5 x 10^5 N/C

8 0
3 years ago
The body went 450 meters within 30 seconds of starting the movement. In what time did the first 50 meters go?
Andrew [12]

3.33 seconds.

<u>Explanation:</u>

We can find the speed of the body using the formula,

Speed = Distance traveled in meters /  time taken in seconds

= 450 m / 30 seconds

= 15 m/s

So per second the distance traveled by the body is 15 m.

So time needed to travel 50 m can be found as,

time = distance/speed

= 50 m / 15 m /s

= 3.33 s

8 0
3 years ago
What is the scientific definition of tracked?
dexar [7]
To find or discover by investigation?
8 0
3 years ago
A block with mass m = 0.450 kg is attached to one end of an ideal spring and moves on a horizontal frictionless surface. The oth
svetoff [14.1K]

Answer:

k = 26.25 N/m

Explanation:

given,

mass of the block= 0.450

distance of the block = + 0.240

acceleration = a_x = -14.0 m/s²

velocity = v_x = + 4 m/s

spring force constant (k) = ?

we know,

x = A cos (ωt - ∅).....(1)

v = - ω A cos (ωt - ∅)....(2)

a = ω²A cos (ωt - ∅).........(3)

\omega = \sqrt{\dfrac{k}{m}}

now from equation (3)

a_x = \dfrac{k}{m}x

k = \dfrac{m a_x}{x}

k = \dfrac{0.45 \times (-14)}{0.24}

k = 26.25 N/m

hence, spring force constant is equal to k = 26.25 N/m

8 0
3 years ago
Read 2 more answers
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