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3 years ago

A football player who weighs 550 N stands indoors wearing her football boots. The boot’s

Physics

1 answer:

EleoNora [17]3 years ago

3 0

Answer:

183333 Pa

Explanation:

The weight of the football player is : 550 N ,thus the force the player exerts on the floor is 550 N

The area of blades in contact with the floor is = 30cm² = 0.003 m²

Pressure = Force / Area

Pressure = 550 / 0.003

Pressure = 183333 Pa

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OLga [1]

Answer:

ggn

Explanation:

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3 years ago

A river flows due east at 1.70 m/s. A boat crosses the river from the south shore to the north shore by maintaining a constant v

ad-work [718]

Answer:

v = 14.1028 m/s

∅ = 83.0765° north of east

the required distance is 40.98 m

Explanation:

Given that;

velocity of the river u = 1.70 m/s

velocity of boat v = 14.0 m/s

Now to get the velocity of the boat relative to shore;

( north of east), we say

a² + b² = c²

(1.70)² + (14.0)² = c²

2.89 + 196 = c²

198.89 = c²

c = √198.89

c = 14.1028 m/s

tan∅ = v/u = 14 / 1.7 = 8.23529

∅ = tan⁻¹ ( 8.23529 ) = 83.0765° north of east

Therefore, the velocity of the boat relative to shore is;

v = 14.1028 m/s

∅ = 83.0765° north of east

width of river = 340 m,

ow far downstream has the boat moved by the time it reaches the north shore in meters = ?

we say;

340sin( 90° - 83.0765°)

⇒ 340sin( 6.9235°)

= 40.98 m

Therefore, the required distance is 40.98 m

5 0

3 years ago

A particle with mass 1.81×10−3 kg and a charge of 1.22×10−8 C has, at a given instant, a velocity v⃗ =(3.00×104m/s)j^. What are

slava [35]

Answer:

The magnitude and direction of the acceleration of the particle is $a= 0.3296\ \hat{k}\ m/s^2$

Explanation:

Given that,

Mass $m = 1.81\times10^{-3}\ kg$

Velocity $v = (3.00\times10^{4}\ m/s)j$

Charge $q = 1.22\times10^{-8}\ C$

Magnetic field $B= (1.63\hat{i}+0.980\hat{j})\ T$

We need to calculate the acceleration of the particle

Formula of the acceleration is defined as

$F = ma=q(v\times B)$

$a =\dfrac{q(v\times B)}{m}$

We need to calculate the value of $v\times B$

$v\times B=(3.00\times10^{4}\ m/s)j\times(1.63\hat{i}+0.980\hat{j})$

$v\times B=4.89\times10^{4}$

Now, put the all values into the acceleration 's formula

$a =\dfrac{1.22\times10^{-8}\times(-4.89\times10^{4}\hat{k})}{1.81\times10^{-3}}$

$a= -0.3296\ \hat{k}\ m/s^2$

Negative sign shows the opposite direction.

Hence, The magnitude and direction of the acceleration of the particle is $a= 0.3296\ \hat{k}\ m/s^2$

7 0

3 years ago

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What is the kinetic energy of a 478 kg object that is moving with a speed of 15 m/s

Tpy6a [65]

The kinetic energy would be 53,775J:)

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3 years ago

What is the mass of a child in a wagon that has a velocity of 10 m/s and a Momentum of 30 KG* M/S

LenKa [72]

Explanation:

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then : 30 = 10 × m and m = 30 ÷ 10 = 3 kg

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3 years ago