All categories

2 years ago

A 10N force pulls to the right and friction opposes 2N. If the object is 20kg,find the acceleraton.

Physics

2 answers:

zmey [24]2 years ago

8 0

Force = mass * acceleration

10 N - 2 N = 20 kg * acceleration

8 N = 20 kg * acceleration

8 / 20 = acceleration

2/5 m/s^2 = acceleration

tia_tia [17]2 years ago

7 0

Answer:

The acceleration is 0.4 m/s²

Explanation:

Given that,

Force = 10 N

Opposing force = 2 N

Mass of object = 20 kg

We need to calculate the acceleration

Using formula of newton's second law

$F = ma$

Where, F = force

m = mass

a = acceleration

Put the value into the formula

$10N-2N = 20\times a$

$8 N=20\times a$

$a =\dfrac{8}{20}$

$a=0.4\ m/s^2$

Hence, The acceleration is 0.4 m/s²

You might be interested in

A rectangular copper strip 1.5cm wide and 0.10cn thick carries a current of 5.0A. Find the Hall voltage for a 1.2T magnetic fiel

Scrat [10]

Answer:

4.4345× 10^-7V

Explanation:

The computation of the half voltage for a 1.2T magnetic field applied is shown below

The volume of one mole of copper is

v = m ÷p

= 63.5 ÷ 8.92

= 7.12cm

Now the density of free electrons in copper is

n = Na ÷ V

= 6.02 × 10^23 ÷ 7.12

= 8.456× 10^28/m^3

Now the half voltage is

= IB ÷ nqt

= (5 × 1.20) ÷ (8.456× 10^28 × 1.6 × 10^-19 × 0.1× 10^-2)

= 4.4345× 10^-7V

7 0

2 years ago

A 0.89 m aqueous solution of an ionic compound with the formula mx has a freezing point of -3.0 ∘c . van't hoff factor?

BigorU [14]

Answer is: Van't Hoff factor (i) for this solution is 1,81.
Change in freezing point from pure solvent to solution: ΔT =i · Kf · b.
Kf - molal freezing-point depression constant for water is 1,86°C/m.
b - molality, moles of solute per kilogram of solvent.
b = 0,89 m.
ΔT = 3°C = 3 K.
i = 3°C ÷ (1,86 °C/m · 0,89 m).

i = 1,81.

5 0

3 years ago

A train traveling at 25 m/s is blowing its whistle at 440 Hz as it crosses a level crossing. You are waiting at the crossing and

ohaa [14]

Answer:

b) 472HZ, 408HZ

Explanation:

To find the frequencies perceived when the bus approaches and the train departs, you use the Doppler's effect formula for both cases:

$f_o=f\frac{v_s+v_o}{v_s-v}\\\\f_o=f'\frac{v_s-v_o}{v_s+v}\\\\$

fo: frequency of the source = 440Hz

vs: speed of sound = 343m/s

vo: speed of the observer = 0m/s (at rest)

v: sped of the train

f: frequency perceived when the train leaves us.

f': frequency when the train is getTing closer.

Thus, by doing f and f' the subjects of the formulas and replacing the values of v, vo, vs and fo you obtain:

$f=f_o\frac{v_s-v}{v_s+v_o}=(440Hz)\frac{340m/s-25m/s}{340m/s}=408Hz\\\\f'=f_o\frac{v_s+v}{v_s-v_o}=(440Hz)\frac{340m/s+25m/s}{340m/s}=472Hz$

hence, the frequencies for before and after tha train has past are

b) 472HZ, 408HZ

6 0

2 years ago

Stephanie serves a volleyball from a height of 0.80 m and gives it an initial velocity of +7.2 m/s straight up. how high will th

Papessa [141]

3.78 m Ignoring resistance, the ball will travel upwards until it's velocity is 0 m/s. So we'll first calculate how many seconds that takes. 7.2 m/s / 9.81 m/s^2 = 0.77945 s The distance traveled is given by the formula d = 1/2 AT^2, so substitute the known value for A and T, giving d = 1/2 A T^2 d = 1/2 9.81 m/s^2 (0.77945 s)^2 d = 4.905 m/s^2 0.607542 s^2 d = 2.979995 m So the volleyball will travel 2.979995 meters straight up from the point upon which it was launched. So we need to add the 0.80 meters initial height. d = 2.979995 m + 0.8 m = 3.779995 m Rounding to 2 decimal places gives us 3.78 m

7 0

2 years ago

Which of the following is not a characteristic of S waves?

9966 [12]

Number a is a correct one

7 0

2 years ago