Ask question

Ask question

All categories

adelina 88 [10]

3 years ago

5

A .2.-kg soccer ball is rolling at 6.0 m/s toward a player. The player kicks the ball back and gives it a velocity of 14 m/s in

the opposite direction.

1 answer:

Sergeu [11.5K]3 years ago

5 0

Answer:

25(6+14) = 5

25(6+14) = 55/.02 = 500/2 = 250 N

25(6+14) = 55/.02 = 500/2 = 250 Na = F/m = 250 N/.25 = 1000 m/s^2

You might be interested in

A rectangular coil of dimensions 5.40cm x 8.50cm consists of25 turns of wire. The coil carries a current of 15.0 mA.

Kazeer [188]

Answer:

(a) Magnetic moment will be $17.212\times 10^{-4}A-m^2$

(b) Torque will be $6.024\times 10^{-4}N-m$

Explanation:

We have given dimension of the rectangular 5.4 cm × 8.5 cm

So area of the rectangular coil $A=5.4\times 8.5=45.9cm^2=45.9\times 10^{-4}m^2$

Current is given as $i=15mA=15\times 10^{-3}A$

Number of turns N = 25

(A) We know that magnetic moment is given by $magnetic\ moment=NiA=25\times 45.9\times 10^{-4}\times 15\times 10^{-3}=17.212\times 10^{-4}A-m^2$

(b) Magnetic field is given as B = 0.350 T

We know that torque is given by $\tau =BINA=0.350\times 15\times 10^{-3}\times 25\times 45.9\times 10^{-4}=6.024\times 10^{-4}N-m$

4 0

3 years ago

A car enters a level, unbanked semi-circular hairpin turn of 100 m radius at a speed of 28 m/s. The coefficient of friction betw

meriva

Answer:

As 28m/s = 28m/s

Explanation:

r = the radius of the curve

m = the mass of the car

μ = the coefficient of kinetic friction

N = normal reaction

When rounding the curve, the centripetal acceleration is

$a = \frac{v^{2}}{r}$

therefore

$\mu mg = m \frac{v^{2}}{r} \\\\ \mu = \frac{v^{2}}{rg}$

$v = \sqrt{\mu rg}$

$\mu = \sqrt{0.8 \times 100\times9.8} \\\\= 28m/s$

As 28m/s = 28m/s

8 0

3 years ago

A force on a particle depends on position such that F(x) = (3.00 N/m2)x2 + (6.00 N/m)x for a particle constrained to move along

Pachacha [2.7K]

Answer:

The work done by a particle from x = 0 to x = 2 m is 20 J.

Explanation:

A force on a particle depends on position constrained to move along the x-axis, is given by,

$F(x)=(3\ N/m^2)x^2+(6\ N/m)x$

We need to find the work done on a particle that moves from x = 0.00 m to x = 2.00 m.

We know that the work done by a particle is given by the formula as follows :

$W=\int\limits {F{\cdot} dx}$

$W=\int\limits^2_0 {(3x^2+6x){\cdot} dx} \\\\W={(x^3}+3x^2)_0^2\\\\\W={(2^3}+3(2)^2)\\\\W=20\ J$

So, the work done by a particle from x = 0 to x = 2 m is 20 J. Hence, this is the required solution.

3 0

3 years ago

Read 2 more answers

The mass of a basketball is three times greater than the mass of a softball . Compare the momentum’s of a softball and a basketb

Lemur [1.5K]

They are moving the softball of basketball

4 0

3 years ago

A sailfish swims 120 km/hr. How far will it travel in 8.0 minutes?

Rasek [7]

Answer:

16km

Explanation:

First change the minutes into hours then multiply by the distance.

(8÷60)×120=16km

5 0

4 years ago