Ask question

Ask question

All categories

2 years ago

13

A horizontal force of 24 N is being applied to a box on a level surface. The box is accelerating at the rate of 2.0 against a fr

ictional force of 14 N. What is the mass of the box?

1 answer:

Leya [2.2K]2 years ago

7 0

Answer:

For SI units the mass m = 5 kg.

Explanation:

The sum of all horizontal forces is must be equal to maₓ:

∑Fₓ = maₓ = 24N - 14N = 10N

maₓ = 10N

m = 10N / aₓ = 10N / 2 m/s² = 5 m/s

You might be interested in

1) Air modeled as an ideal gas enters a well-insulated diffuser operating at steady state at 270 K with a velocity of 180 m/s an

GarryVolchara [31]

Answer:

Explanation:

Kinetic energy of a gas molecule = 3/2 k T

1/2 m v² = 3/2 k T

v² ∝ T

(v₁ / v₂)² = T₁ / T₂

( 180 / 48.4 )² = 270 / T₂

13.83 = 270 / T₂

T₂ = 19.52 K .

Exit temp = 19.52 K .

5 0

2 years ago

A dog goes for a walk in the neighborhood leaves home and take several displacements as follows. It first head

Korvikt [17]

Answer:

sorry i couldn't answer your question I tried

4 0

1 year ago

A 225-kg object and a 525-kg object are separated by 3.80 m. (a) Find the magnitude of the net gravitational force exerted by th

pav-90 [236]

Answer: $F_{net}=3.383\times 10^{-7}\ N$

Explanation:

Given

Mass of first object $m_1=225\ kg$

Mass of second object $m_2=525\ kg$

Distance between them $d=3.8\ m$

$m_3=61\ kg$ object is placed between them

So force exerted by $m_1$ on $m_3$

$F_{13}=\frac{Gm_1m_3}{1.9^2}$

$F_{13}=\frac{6.674\times 10^{-11}(225\times 61)}{1.9^2}$

$F_{13}=2.5374141274×10^{−7}\ N$

Force exerted by $m_2\ on\ m_3$

$F_{23}=\frac{Gm_2m_3}{1.9^2}$

$F_{23}=\frac{6.674\times 10^{-11}(525\times 61)}{1.9^2}$

$F_{23}=5.920632964\times 10^{-7}\ N$

So net force on $m_3$ is

$F_{net}=F_{23}-F_{13}$

$F_{net}=5.920632964\times 10^{-7}-2.5374141274\times 10^{-7}$

$F_{net}=3.383\times 10^{-7}\ N$

i.e. net force is towards $m_2$

(b)For net force to be zero on $m_3$ , suppose

So force exerted by $m_1$ and $m_2$ must be equal

$F_{13}=F_{23}$

$\Rightarrow \frac{Gm_1m_3}{x^2}=\frac{Gm_2m_3}{(3.8-x)^2}$

$\Rightarrow \frac{m_1}{x^2}=\frac{m_2}{(3.8-x)^2}$

$\Rightarrow (\frac{3.8-x}{x})^2=\frac{m_2}{m_1}$

$\Rightarrow \frac{3.8-x}{x}=\sqrt{\frac{525}{225}}$

$\Rightarrow 3.8-x=1.52752x$

$\Rightarrow 3.8=2.52x$

$\Rightarrow x=1.507\ m$

4 0

2 years ago

A ball is tied to the end of a cable of negligible mass. The ball is spun in a circle with a radius 2. 00 m making 7. 00 revolut

mr Goodwill [35]

Answer:

F = M ω^2 R is the centripetal force on the ball

f = 7 / 10 = .70 / sec revolutions / sec

ω = 2 pi f = 4.40 / sec

ah = F / m = 4.4^2 * 2 = 38.7 m/s^2

horizontal acceleration = 38.7 m/s^2

vertical acceleration = 9.80 m/s^2

a = (38.7^2 + 9.8^2)^1/2 = 39.9 m/s^2 total acceleration

8 0

1 year ago

4.) An apartment building is on fire and a guy is trapped on the fire escape ladder. There is a

Tems11 [23]

Answer:

5.3 m/s

Explanation:

First, find the time it takes for him to fall 7m.

y = y₀ + v₀ t + ½ at²

0 = 7 + (0) t + ½ (-9.8) t²

0 = 7 − 4.9 t²

t ≈ 1.20 s

Now find the velocity he needs to travel 6.3m in that time.

x = x₀ + v₀ t + ½ at²

6.3 = 0 + v₀ (1.20) + ½ (0) (1.20)²

v₀ ≈ 5.27 m/s

Rounded to two significant figures, the man must run with a speed of 5.3 m/s.

3 0

2 years ago