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natulia [17]
2 years ago
11

A 1,300 kg wrecking ball hits the building at 1.07 m/s2.

Physics
2 answers:
dezoksy [38]2 years ago
7 0
So we know F=ma
m means mass and a means acceleration
so Force= ma
so F=1300X1.07=1391N
hope you like it
if it helps you please mark the answer as brainliest answer
tatuchka [14]2 years ago
6 0

F (force) = mass × acceleration.

<em>The unit N (Newton) is known as kilogram-meter/seconds2. </em>

<u>Thus, F = 1300 kg × 1.07 m/s2 = 1391 N. </u>

<u>The answer is 1391 N.</u>

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A projectile is launched vertically from the surface of the Moon with an initial speed of 1360 m/s. At what altitude is the proj
LenaWriter [7]

Answer:

485520 m

Explanation:

v_{o} = initial velocity of the projectile = 1360 m/s

v_{f} = final velocity of the projectile = \left ( \frac{2}{5} \right )v_{_{o}} = \left ( \frac{2}{5} \right )(1360) = 544 m/s

a = acceleraton due to gravity on moon = - 1.6 m/s²

h = Altitude of the projectile

Using the kinematics equation

v_{f}^{2} = v_{o}^{2} + 2 a h

Inserting the values

544^{2} = 1360^{2} + 2 (-1.6) h

h = 485520 m

3 0
3 years ago
Two protons (each with q = 1.60 x 10-19)
otez555 [7]

Answer:

230.4 N

Explanation:

From the question given above, the following data were obtained:

Charge (q) of each protons = 1.6×10¯¹⁹ C

Distance apart (r) = 1×10¯¹⁵ m

Force (F) =?

NOTE: Electric constant (K) = 9×10⁹ Nm²/C²

The force exerted can be obtained as follow:

F = Kq₁q₂ / r²

F = 9×10⁹ × (1.6×10¯¹⁹)² / (1×10¯¹⁵)²

F = 9×10⁹ × 2.56×10¯³⁸ / 1×10¯³⁰

F = 2.304×10¯²⁸ / 1×10¯³⁰

F = 230.4 N

Therefore, the force exerted is 230.4 N

5 0
2 years ago
URGENTE. ¿Qué fuerzas actúan en un tobogán?
professor190 [17]

Answer:

friction or fricción

if u only speak spanish

Explanation:

5 0
3 years ago
Twist-on connectors without the spring-steel coils (plastic threads only) are suitable for making branch-circuit connections.
Gnoma [55]

Answer:

if it is a plastic connector it wont work but if there is metal or steel it will work

Explanation:

6 0
3 years ago
An early submersible craft for deep-sea exploration was raised and lowered by a cable from a ship. When the craft was stationary
Talja [164]

Answer:

A) 5.2 x 10³ N

B) 8.8 x 10³ N

Explanation:

Part A)

F_{g} = weight of the craft in downward direction = tension force in the cable when stationary = 7000 N

T = Tension force in upward direction

F_{d} = Drag force in upward direction = 1800 N

Force equation for the motion of craft is given as

F_{g} - F_{d} - T = 0

7000 - 1800 - T = 0

T = 5200 N

T = 5.2 x 10³ N

Part B)

F_{g} = weight of the craft in downward direction = tension force in the cable when stationary = 7000 N

T = Tension force in upward direction

F_{d} = Drag force in downward direction = 1800 N

Force equation for the motion of craft is given as

T  - F_{g} - F_{d} = 0

T - 7000 - 1800  = 0

T = 8800 N

T = 8.8 x 10³ N

4 0
3 years ago
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