Help asap please !!!! :(

an 0.45-kg object is being pulled with a tension force of 14.2 n at a constant velocity of 3.41 m/s. what is the force of fricti

Lisa [10]

The forces of friction acting on the object are 12.67 N.

Friction is the force resisting the relative motion of sturdy surfaces, fluid layers, and material factors sliding towards each specific. There are numerous styles of friction. Dry friction is a strain that opposes the relative lateral movement of two stable surfaces in touch.

Calculation:-

ma - 14.2 N = Friction force

14.2 N - 0.45 × 3.41 = 12.67 N

Friction is the shape of contact pressure. It exists between the surfaces that are in touch. The frictional force relies upon the character of the floor in contact. The rougher the ground, the extra friction is involved. The frictional force is proportional to the urgent pressure, which is the burden of the body. it's miles impartial of the place of the touch.

Learn more about friction here:-brainly.com/question/24338873

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7 0

1 year ago

La masa de un objeto en la tierra es de 100 kg.

andreev551 [17]

Answer:

a) 980 N

b) 100 kg

c) 163.33 N

Explanation:

a)

The weight of the object on Earth is the product of its mass times the acceleration due to gravity, that is:

$W=m\,g=(100\,\,kg)\,g=(100\,\,kg)\,(9.8\,\,m/s^2)=980\,\,N$

b)

The mass of the object on the moon, is exactly the same (100 kg) since mass is an inherited value of the object.

c)

Considering that the acceleration of gravity on the Moon is about 1/6 that we have on earth (g), then the weight of the object on the Moon becomes:

$W=m\,\frac{g}{6} =\frac{980}{6}\,\,N=163.33\,\,N$

3 0

3 years ago

You are climbing in the high sierra when you suddenly find yourself at the edge of a fog-shrouded cliff. to find the height of t

sergey [27]

The total time it takes the sound to reach our on top of the cliff (10.0 s) is actually sum of two times:
- the time it takes the rock to hit the ground starting from the top of the cliff, t1
- the time it takes the sound to reach the top of the cliff starting from the ground, t2
1) The rock moves by uniformly accelerated motion, with constant acceleration $g=9.81 m/s^2$ . Its law of motion is given by $y(t)=h-v_0t-\frac{1}{2}gt^2$ , where $v_0=0$ is the initial velocity of the rock, and h is the height of the cliff. The time t1 is the time at which the rock reaches the ground, so that y(t1)=0, and the equation becomes
$0=h-\frac{1}{2}gt_1^2$
2) The sound moves from the ground to the top of the cliff by uniform motion, with constant speed v=343 m/s. Therefore, the sound covers the distance h (the height of the cliff) in a time t2 given by
$h=vt_2$
3) If we rewrite h in both equations, we can write:
$=\frac{1}{2}gt_1^2=vt_2$ (1)
4) We also know that the sum of t1 and t2 is equal to 10 seconds:
$t_1+t_2=10$
from which we find
$t_2=10-t_1$
if we substitute this into eq.(1), we get
$\frac{1}{2}gt_1^2+vt_1-10v=0$
Numerically:
$4.9t_1^2 + 343 t_1 - 3430=0$
Solving the equation, we find the solution $t_1=8.87 s$ (the other solution is negative, so it does not have physical meaning). As a consequence,
$t_2 = 10-t_1 = 1.13 s$
and the height of the cliff is given by
 $h=vt_2=(343 m/s)(1.13 s)=388 m$

3 0

3 years ago

Read 2 more answers

A longitudinal wave is a combination of a transverse wave and a surface wave.(true or false)

Lisa [10]

Answer: True

Explanation:

8 0

3 years ago

A tree falls in a forest. How many years must pass before the 14C activity in 1.03 g of the tree's carbon drops to 1.02 decay pe

Illusion [34]

Answer:

t = 5.59x10⁴ y

Explanation:

To calculate the time for the ¹⁴C drops to 1.02 decays/h, we need to use the next equation:

$A_{t} = A_{0}\cdot e^{- \lambda t}$ (1)

where $A_{t}$ : is the number of decays with time, A₀: is the initial activity, λ: is the decay constant and t: is the time.

To find A₀ we can use the following equation:

$A_{0} = N_{0} \lambda$ (2)

where N₀: is the initial number of particles of ¹⁴C in the 1.03g of the trees carbon

From equation (2), the N₀ of the ¹⁴C in the trees carbon can be calculated as follows:

$N_{0} = \frac{m_{T} \cdot N_{A} \cdot abundance}{m_{^{12}C}}$

where $m_{T}$ : is the tree's carbon mass, $N_{A}$ : is the Avogadro's number and $m_{^{12}C}$ : is the ¹²C mass.

$N_{0} = \frac{1.03g \cdot 6.022\cdot 10^{23} \cdot 1.3\cdot 10^{-12}}{12} = 6.72 \cdot 10^{10} atoms ^{14}C$

Similarly, from equation (2) λ is:

$\lambda = \frac{Ln(2)}{t_{1/2}}$

where t 1/2: is the half-life of ¹⁴C= 5700 years

$\lambda = \frac{Ln(2)}{5700y} = 1.22 \cdot 10^{-4} y^{-1}$

So, the initial activity A₀ is:

$A_{0} = 6.72 \cdot 10^{10} \cdot 1.22 \cdot 10^{-4} = 8.20 \cdot 10^{6} decays/y$

Finally, we can calculate the time from equation (1):

$t = - \frac{Ln(A_{t}/A_{0})}{\lambda} = - \frac {Ln(\frac{1.02decays \cdot 24h \cdot 365d}{1h\cdot 1d \cdot 1y \cdot 8.20 \cdot 10^{6} decays/y})}{1.22 \cdot 10^{-4} y^{-1}} = 5.59 \cdot 10^{4} y$

I hope it helps you!

4 0

3 years ago