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babunello [35]
2 years ago
9

An object of mass 25 kg acted upon by a net force of 10 N will experience an acceleration of?

Physics
1 answer:
Zolol [24]2 years ago
6 0

Answer:

0.4 m/s2

Explanation:

mass: 25kg

net force: 10 N

acceleration: ?

net force ÷ by mass= acceleration

10 N ÷ 25 Kg = 0.4 m/s2

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Two forces act on a 41-kg object. One force has magnitude 65 N directed 59° clockwise from the positive x-axis, and the other ha
Genrish500 [490]

Answer:

a = 1.41 m/s²

Explanation:

Given that

mass ,m= 41 kg

F₁ = 65 N , θ = 59°

F₂ = 35 N ,θ = 32°

The component of Force F₁

F₁x= F₁cos59° i

F₁x= 65 x cos59° i = 33.47 i

F₁y= - F₁ sin 59° j

F₁y= - 65 x sin 59° j = - 55.71 j

The component of Force F₂

F₂x= F₂ sin 32° i

F₂x= 35 x sin 32° i = 18.54 i

F₂y=  F₂ cos 32° j

F₂y=  35 x cos 32° j =  29.68 j

The total force F

F= 33.47 i +   18.54 i - 55.71 j +   29.68 j

F= 52.01 i - 26.03 j

The magnitude of the force F

F=\sqrt{52.01^2+26.03 ^2}\ N

F=58.16 N

We know that

F= m a

a= Acceleration

m=mass

58.16 = 41 x a

a = 1.41 m/s²

6 0
3 years ago
A cube with sides of area 18 cm^2 contains a 6.0 nanoCoulomb charge. Find the flux of the electric field through the surface of
Vanyuwa [196]

Answer:

The flux of the electric field  is 677.6 Nm²/C

Explanation:

Given that,

Area = 18 cm²

Charge = 6.0 nC

We need to calculate the flux of the electric field

Using Gauss's law

\phi=\dfrac{q}{\epsilon_{0}}

Where, q = charge

\epsilon_{0} =permittivity of free space

Put the value into the formula

\phi=\dfrac{6.0\times10^{-9}}{8.854\times10^{-12}}

\phi=677.6\ Nm^2/C

Hence, The flux of the electric field  is 677.6 Nm²/C.

4 0
3 years ago
A train starts from rest with acceleration of 0.5ms-2 find speed in Kmh-1 when it moved through 100m. (36kmh-1)
SashulF [63]

Answer:

36km/h

Explanation:

since the acceleration of the train was 0.5m/s² i.e it is constant . since the acceleration is constant you can use one of the "SUVAT" equations to find the answer.

the equation that best fits this question is: v²=u² + 2āS.

v stands for final velocity

u stands for intial velocity

v²=u² + 2āS since the car started from rest the intial velocity is zero.

v²=u² + 2āS

v²= 0 + 2(0.5m/s²)*100m

v²= 1m/s²*100m

√v²=√100m²/s²

v=10m/s

1km/hr = 1/3.6m/s

x. = 10m/s.

use criss cross method to find the value of x

x=<u>( 1km/hr* 10m/s)</u>

1/3.6m/s

x= 36km/hr

4 0
3 years ago
To which abiotic factor is the biotic factor in this photograph adapted?
kompoz [17]

the pickle

because it is living, I'm assuming

5 0
3 years ago
The average marathon runner can complete the 42.2-km distance of the marathon in 3 h and 30 min. If the runner's mass is 85 kg,
Semenov [28]

Answer:

the runner's average kinetic energy during the run is 476.96 J.

Explanation:

Given;

mass of the runner, m = 85 kg

distance covered by the runner, d = 42.2 km = 42,200 m

time to complete the race, t = 3 hours 30 mins = (3 x 3600s) + (30 x 60s)

                                                                               = 12,600 s

The speed of the runner, v = d/t

                                          v = 42,200 / 12,600

                                          v = 3.35 m/s

The runner's average kinetic energy during the run is calculated as;

K.E = ¹/₂mv²

K.E = ¹/₂ × 85 × (3.35)²

K.E = 476.96 J

Therefore, the runner's average kinetic energy during the run is 476.96 J.

5 0
2 years ago
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