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3 years ago

13

A horizontal force of 90.0 N is required to push a 75kg object along a horizontal surface at a constant speed. What is the magni

tude of the force of friction. Please show your work I am having troubles.

1 answer:

Varvara68 [4.7K]3 years ago

6 0

If the object is moving at a constant speed, acceleration is 0. So:

$F = ma

F = m\times 0

F = 0$

The resultant force is 0. So the force pushing the object must be equal to the friction force acting, so

Friction = 90 N

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Susan's 12.0 kg baby brother Paul sits on a mat. Susan pulls the mat across the floor using a rope that is angled 30∘ above the

Pavlova-9 [17]

Answer:1.71 m/s

Explanation:

Given

mass of Susan $m=12 kg$

Inclination $\theta =30^{\circ}$

Tension $T=29 N$

coefficient of Friction $\mu =0.18$

Resolving Forces Along x axis

$F_x=T\cos \theta -f_r$

where $f_r=friction\ Force$

$F_y=mg-N-T\sin \theta$

since there is no movement in Y direction therefore

$N=mg-T\sin \theta$

and $f_r=\mu N$

Thus $F_x=T\cos \theta -\mu N$

$F_x=29\cos (30)-\0.18\times (12\times 9.8-29\sin (30))$

$F_x=25.114-18.558$

$F_x=6.556 N$

Work done by applied Force is equal to change to kinetic Energy

$F_x\cdot x=\frac{1}{2}\cdot mv_f^2-\frac{1}{2}\cdot mv_i^2$

$6.556\times 2.7=\frac{1}{2}\cdot 12\times v_f^2$

$v_f^2=\frac{6.556\times 2.7\times 2}{12}$

$v_f^2=2.95$

$v_f=1.717 m/s$

8 0

2 years ago

Symptoms of excessive stress include all of the following EXCEPT:

sweet-ann [11.9K]

Answer:

Symptoms of excessive stress include all of the following EXCEPT: increased energy level.

6 0

2 years ago

Consider one such cell where the magnitude of the potential difference is 65 mV, and the inner surface of the membrane is at a h

Gelneren [198K]

Answer: W = $1.04.10^{-20}$ J

Explanation: Since the potassium ion is at the outside membrane of a cell and the potential here is lower than the potential inside the cell, the transport will need work to happen.

The work to transport an ion from a lower potential side to a higher potential side is calculated by

$W=q.\Delta V$

q is charge;

ΔV is the potential difference;

Potassium ion has +1 charge, which means:

p = $1.6.10^{-19}$ C

To determine work in joules, potential has to be in Volts, so:

$\Delta V=65.10^{-3}V$

Then, work is

$W=1.6.10^{-19}.65.10^{-3}$

$W=1.04.10^{-20}$

To move a potassium ion from the exterior to the interior of the cell, it is required $W=1.04.10^{-20}$ J of energy.

8 0

3 years ago

A rocket initially at rest accelerates at a rate of 99. 0 meters/second2. Calculate the distance covered by the rocket if it att

creativ13 [48]

The rocket will cover $1 \times 10^3\rm \ m$ distance in 4. 5 s. Acceleration can be defined as the change in velocity.

<h3>What is acceleration?</h3>

Acceleration can be defined as the change in speed or the direction of the object.

From kinamatic equation:

$D = v_{t} +\dfrac 12at^2$

Where,

$D$ - final velocity = 445 m/s

$v_0$ - initial valocity = 0 m/s

$a$ - acceleration = 99. 0 m/s²

$t$ - time = 4. 50 s

Put the values in the formula,

$D = 0\times ( 4.5) + \dfrac12\times (99)(4.5)^2\\\\D = 1002 {\rm \ m}\\\\D = 1 \times 10^3\rm \ m$

Therefore, the rocket will cover $1 \times 10^3\rm \ m$ distance in 4. 5 s.

Learn more about Acceleration :

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2 years ago

A singly charged positive ion has a mass of 3.46 × 10−26 kg. After being accelerated through a potential difference of 215 V the

jasenka [17]

Answer:

1.8 cm

Explanation:

$m$ = mass of the singly charged positive ion = 3.46 x 10⁻²⁶ kg

$q$ = charge on the singly charged positive ion = 1.6 x 10⁻¹⁹ C

$\Delta V$ =Potential difference through which the ion is accelerated = 215 V

$v$ = Speed of the ion

Using conservation of energy

Kinetic energy gained by ion = Electric potential energy lost

$(0.5) m v^{2} = q \Delta V\\(0.5) (3.46\times10^{-26}) v^{2} = (1.6\times10^{-19}) (215)\\(1.73\times10^{-26}) v^{2} = 344\times10^{-19}\\v = 4.5\times10^{4} ms^{-1}$

$r$ = Radius of the path followed by ion

$B$ = Magnitude of magnetic field = 0.522 T

the magnetic force on the ion provides the necessary centripetal force, hence

$qvB = \frac{mv^{2} }{r} \\qB = \frac{mv}{r}\\r =\frac{mv}{qB}\\r =\frac{(3.46\times10^{-26})(4.5\times10^{4})}{(1.6\times10^{-19})(0.522)}\\r = 0.018 m \\r = 1.8 cm$

5 0

2 years ago