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evablogger [386]
3 years ago
11

Explain the conditions through which friction can be increased.​

Physics
2 answers:
Rufina [12.5K]3 years ago
8 0

Answer:

Friction can be increased the longer two objects are rubbed together, since the longer they are rubbed is the more heat produced.

Explanation:

Thepotemich [5.8K]3 years ago
6 0

Answer:

walking, car moving

Explanation:

basically anything with two surfaces pressing together ig

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A large blue marble of mass 3.5 g is moving to the right with a velocity of 15 cm/s. The large marble hits a small red marble of
Aneli [31]

Explanation:

Let's solve the problem by using conservation of momentum:

m1 u1 + m2 u2 = m1 v1 + m2 v2

where:

m1 = 3.5 g  is the mass of the blue marble

m2= 1.2 g is the mass of the red marble

u1 = 15 cm/s  is the initial velocity of the blue marble

u2 = 3.5 cm/s is the initial velocity of the red marble

v1 = 5.5 cm/s is the final velocity of the blue marble

We can find the final velocity of the red marble by re-arranging the equation and solving for v2 :

v2 = 1/m²(m1 u1 + m2 u2 - m1 v1)=

=1/1.2(3.5×15+1.2×3.5−3.5×5.5 )=31cm/s

7 0
4 years ago
Encontrar el peso de un elefante cuya masa es de 4500kg. Recordar que la gravedad es de 9.8m/s2
dusya [7]

la respuesta es 459.183673 kg s2 / m

7 0
3 years ago
Read 2 more answers
Electricity The power P, in watts, that a circular solar cell produces and radius of the cell in centimeters are related by the
gavmur [86]

Answer:

9.05 W

Explanation:

The given formula is r=\sqrt {\frac {P}{0.02\pi}} where r is in centimeters and P is in Watts

Making Power, P the subject from the above formula

P=0.02\pi r^{2}

Substituting r with 12 cm then

P=0.02\pi 12^{2}=
9.047786842  W\approx 9.05 W

5 0
3 years ago
A cube 10.0 cm on each side has a density of 2.053 x 10^4 kg/m^3. It’s apparent weight in fresh water is 192 N. Find the buoyant
insens350 [35]

Answer:

9 N

Explanation: Given that L = side of the cube = 10cm = 0.1m

Mass = Density * Volume

Let’s determine the volume of the cube in m^3.

V = L^3

V = 0.1^3 = 0.001m^3

Mass = 0.001 * 2.053 * 10^4 = 20.53kg

Weight = 20.53 * 9.8 = 201.194

Buoyant force = 201.194 – 192 = 9.194 N

This is approximately 9 N.

6 0
4 years ago
A flywheel of J = 50 kg-m2 initially standing still is subjected to a constant torque. If the angular velocity reaches 20 Hz in
Karolina [17]

Answer:

\tau = 1256.5\ N.m

Explanation:

given,

J = 50 kg-m²

frequency, f = 20 Hz

time ,t = 5 s

we know,

angular velocity = 2 π f

ω = 2 π x 20

ω = 125.66 rad/s

now, angular acceleration calculation

\alpha = \dfrac{\omega_f-\omega_i}{t}

\alpha = \dfrac{125.66-0}{5}

   α = 25.13 rad/s²

Torque given to the flywheel.

    \tau = I \alpha

    \tau = 50\times 25.13

    \tau = 1256.5\ N.m

Torque of the given  flywheel is equal to \tau = 1256.5\ N.m

7 0
4 years ago
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