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4 years ago

A 50kg girl is playing soccer. She runs towards the ball at a speed of 7 m/s. How much kinetic energy does she have as she runs

Physics

2 answers:

stepladder [879]4 years ago

6 0

Yup its 1,225j thank you lol i need more wordsssssss

stepladder [879]4 years ago

3 0

The calculation for kinetic energy is this

KE = 1/2mv^2

KE = 1/2(50)(7^2)

KE = 1/2(49•50)

KE = 1225 kgm^2/s^2.

Or simply 1225 J.

She possess this much energy when she runs.

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When neither air mass displaces the adjacent one, their boundary is called a(an) ____________ front.

Nataly_w [17]

Answer:

Stationary front

Explanation:

A stationary front is usually defined as a front when both the warm as well as cold front does not move and remain stationary. It is because the force exerted by both the air masses against each other is not so high to displace one another. In this case, the wind blows parallel instead of blowing perpendicular to the front.

Generally, in weather maps, the stationary front is shown by the alternate use of blue triangles and red semi-circles, where the triangle faces the warm front and the semi-circle faces the cold front.

8 0

4 years ago

A circular wire loop of radius 14.1 cm carries a current of 2.34 A. It is placed so that the normal to its plane makes an angle

Tomtit [17]

Answer:

(a) $0.1460Am^2$

(b) 1.3927 Nm

Explanation:

We have given that radius $r=14.1cm=0.141m$

Current i = 2.34 A

Angle $\Theta =49.2^{\circ}$

Uniform magnetic field B = 12.6 T

Area is given by $A=\pi r^2=3.14\times 0.141^2=0.0624m^2$

(A) We know that magnetic dipole moment is given by $M=iA$ , here i is current and A is area

So magnetic moment $M=iA=2.34\times 0.0624=0.1460Am^2$

(b) We know that torque is given by

$\tau =BIAsin\Theta$

So torque experienced by coil will be $\tau =BIAsin\Theta =12.6\times 2.34\times 0.0624\times sin49.2^{\circ}=1.3927Nm$

7 0

3 years ago

The student in item 1 moves the box up a ramp inclined at 12 degrees with the horizontal. If the box starts from rest at the bot

Alik [6]

Answer:

Acceleration up the ramp = -2.745 m/ $s^{2}$

Explanation:

Given

box is pulled at angle Θ = $25^{o}$

Force applied F = 185N

coefficient of friction, μ $_{k}$ = 0.27

mass of the box m = 35 kg

We know that,

acceleration due to gravity g = 9.8 m/ $s^{2}$

horizontal component $F_{x}$ = F cosΘ = 185 * cos $25^{o}$ =167.67

vertical component $F_{y}$ = F sinΘ =185*sin $25^{o}$ = 78.18

Another vertical component is due to gravity $F_{g}$ , force in given by

$F_{g}$ = mg

= 35 x 9.8

= 343.35 N

Normal force $F_{n}$ = $F_{g}$ - $F_{y}$

= 343.35 - 78.18

= 265.17 N

Frictional force $F_{k}$ = $F_{n}$ * μ $_{k}$

= 265.17 * 0.27

= 71.596 N

To find acceleration, we know that,

force = mass x acceleration

acceleration = $\frac{force}{mass}$

Here force is the summation of frictional force and horizontal component of the applied force. These force act in opposite directions.

force = $F_{k}$ - $F_{x}$

= 71.596 - 167.67

= -96.074

acceleration = $\frac{-96.074}{35}$

= -2.745 m/ $s^{2}$

6 0

3 years ago

PLEASEE HELP WILLL GIVE 15+ POINTS AND BRAINLIEST PLEASEEEEEEEEEEEEEEEEEEEEEEE

Dovator [93]

um you can't really get help on this it's mostly you sorry

4 0

3 years ago

A block of wood of mass 300g and density 0.75 g/cm^3 is floating on the surace of a liquid of density 1.1 g/cm^3. What mass of l

Travka [436]

Answer:

The minimum mass of the lead for the combination to submerge is 155 g.

Explanation:

let M be the mass of the wood.

let m be the minimum mass of lead to be added for the combination to submerge.

let ρ1 be the density of the liquid.

let ρ2 be the density of the wood.

let ρ3 be the density of lead.

let g be the gravitational acceleration.

For the combination to submerge, the weight of the wood combined with the weight of lead should at least be equal to the buyant force, that is:

weight of wood and lead = buyant force

g×(M+m) = g×(ρ1)×(M/ρ2 + m/ρ3)

M+m = (ρ1)×(M/ρ2 + m/ρ3)

m - ρ1×m/ρ3 = (ρ1)×(M/ρ2) - M

m(1 - ρ1/ ρ3) = M(ρ1/ρ2 -1)

m = [M(ρ1/ρ2 -1)]/[(1 - ρ1/ ρ3)]

= [(300)(1.1/0.75 -1)]/[(1 - 1.1/ 11.3)]

= 155 g

Therefore, the minimum mass of the lead for the combination to submerge is 155 g.

4 0

4 years ago