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MAXImum [283]
3 years ago
5

The kinetic energy of an object with a mass of 6.8 kg and a velocity of 5.0 m/s is J. (Report the answer to two significant figu

res.)
Physics
2 answers:
abruzzese [7]3 years ago
8 0

Answer:

\boxed{\sf Kinetic \ energy \ (KE) = 85 \ J}

Given:

Mass (m) = 6.8 kg

Speed (v) = 5.0 m/s

To Find:

Kinetic energy (KE)

Explanation:

Formula:

\boxed{ \bold{\sf KE =  \frac{1}{2} m {v}^{2} }}

Substituting values of m & v in the equation:

\sf \implies KE =  \frac{1}{2}  \times 6.8 \times  {5}^{2}

\sf \implies KE = \frac{1}{ \cancel{2}}  \times  \cancel{2} \times 3.4 \times 25

\sf \implies KE =3.4 \times 25

\sf \implies KE = 85 \: J

Angelina_Jolie [31]3 years ago
4 0

Answer:

85J

Explanation:

got right on edg

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In a hydraulic device, the surface area of the small piston is 1 cm2 and the surface area of the large piston is 8 cm2. If 10 N
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An electromagnet is strong enough to lift 200 kg of mass. If you double the number of turns of wire around the core and triple t
VLD [36.1K]

Answer:1200 kg

Explanation:

Given

Electromagnet can lift a mass of  200\ kg

Magnetic force necessary to hold mass is proportional to Magnetic field

and magnetic field (B) is proportional to NIA

where N=no of turns

I=current in the magnet

A=area of cross-section

When we double the no of turns and triple the current then

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I becomes 3 I

therefore magnetic field becomes 6 times

Thus magnetic force becomes 6 times and we can say that electromagnet can hold the 6 times of given weight

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4 0
3 years ago
What resource is used to make fertilizers and other products such as toothpaste and insecticides?
igomit [66]

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Silica

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5 0
2 years ago
A uniform rod is set up so that it can rotate about an axis at perpendicular to one of its ends. The length and mass of the rod
guapka [62]

Answer:

F = 0.179 N

Explanation:

given,

length of the rod = 0.759 m

mass of the rod = 1.03 kg

force = ?

initial angular speed = 0 rad/s

final angular speed = 6.87 rad/s

time interval = 9.95 s

the angular acceleration

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

\alpha = \dfrac{6.87 -0}{9.95}

\alpha =0.690\ rad/s^2

moment inertial of the rod

I =\dfrac{mL^2}{3}

torque produced by the rod τ = I α

torque is also equal to τ = F L

now,

F L =  I α

F L=(\dfrac{mL^2}{3})\times \alpha

F= (\dfrac{1.03\times 0.759}{3})\times 0.690

F = 0.179 N

5 0
3 years ago
Read 2 more answers
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