Can someone help me with this? I will give brainliest! Ridiculous answers will be reported. Thank you.

The modernization of tennis occurred in which<br> country?

Kobotan [32]

Answer:

The International Lawn Tennis Federation, now known simply as the International Tennis Federation, the sport's governing body, was founded in 1913, composed of 13 national tennis associations. The Modern Olympics , growing out of the ancient tradition, resurfaced under the direction of the International Olympic Committee in Athens in 1896.

Explanation:

8 0

3 years ago

A 10-g bullet moving horizontally with a speed of 2.0 km/s strikes and passes through a 4.0-kg block moving with a speed of 4.2

SVEN [57.7K]

Answer:

$K=512J$

Explanation:

Since the surface is frictionless, momentum will be conserved. If the bullet of mass $m_1$ has an initial velocity $v_{1i}$ and a final velocity $v_{1f}$ and the block of mass $m_2$ has an initial velocity $v_{2i}$ and a final velocity $v_{2f}$ then the initial and final momentum of the system will be:

$p_i=m_1v_{1i}+m_2v_{2i}$

$p_f=m_1v_{1f}+m_2v_{2f}$

Since momentum is conserved, $p_i=p_f$ , which means:

$m_1v_{1i}+m_2v_{2i}=m_1v_{1f}+m_2v_{2f}$

We know that the block is brought to rest by the collision, which means $v_{2f}=0m/s$ and leaves us with:

$m_1v_{1i}+m_2v_{2i}=m_1v_{1f}$

which is the same as:

$v_{1f}=\frac{m_1v_{1i}+m_2v_{2i}}{m_1}$

Considering the direction the bullet moves initially as the positive one, and writing in S.I., this gives us:

$v_{1f}=\frac{(0.01kg)(2000m/s)+(4kg)(-4.2m/s)}{0.01kg}=320m/s$

So kinetic energy of the bullet as it emerges from the block will be:

$K=\frac{mv^2}{2}=\frac{(0.01kg)(320m/s)^2}{2}=512J$

6 0

4 years ago

5. Confirm dimensionally that the product (lvp/n) is

Sidana [21]

Answer:

How do you do it?

Explanation:

7 0

3 years ago

Suppose that a particular artillery piece has a range R = 9880 yards . Find its range in miles. Use the facts that 1mile=5280ft

Solnce55 [7]

Answer:

R = 9880 yd * 3 ft/yd / 5280 ft/mi = 5.61 mi

If you do it in steps

R = 9880 yd * 3 ft/yd = 29640 ft

R = 29640 ft / 5280 ft/mi = 5.61 mi

6 0

3 years ago

An object with mass 100 kg moved in outer space. When it was at location <8, -30, -4> its speed was 5.5 m/s. A single cons

Alenkasestr [34]

Answer:

v = ( 6.41 i^ + 8.43 j^ + 2.63 k^ ) m/s

Explanation:

We can solve this problem using the kinematic relations, we have a three-dimensional movement, but we can work as three one-dimensional movements where the only parameter in common is time (a scalar).

X axis.

They indicate the initial position x = 8 m, its initial velocity v₀ = 5.5 m / s, the force Fx₁ = 220 N x₁ = 14 m, now the force changes to Fx₂ = 100 N up to the point xf = 17 m. The final speed is wondered.

As this movement is in three dimensions we must find the projection of the initial velocity in each axis, for this we can use trigonometry

the angle fi is with respect to the in z and the angle theta with respect to the x axis.

sin φ = z / r

Cos φ = r_p / r

z = r sin φ

r_p = r cos φ

the modulus of the vector r can be found with the Pythagorean theorem

r² = (x-x₀) ² + (y-y₀) ² + (z-z₀) ²

r² = (14-8) 2 + (-21 + 30) 2+ (-7 +4) 2

r = √126

r = 11.23 m

Let's find the angle with respect to the z axis (φfi)

φ = sin⁻¹ z / r

φ = sin⁻¹ ( $\frac{-7+4}{11.23}$ )

φ = 15.5º

Let's find the projection of the position vector (r_p)

r_p = r cos φ

r_p = 11.23 cos 15.5

r_p = 10.82 m

This vector is in the xy plane, so we can use trigonometry to find the angle with respect to the x axis.

cos θ = x / r_p

θ = cos⁻¹ x / r_p

θ = cos⁻¹ ( $\frac{14-8}{10.82}$ )

θ = 56.3º

taking the angles we can decompose the initial velocity.

sin φ = v_z / v₀

cos φ = v_p / v₀

v_z = v₀ sin φ

v_z = 5.5 sin 15.5 = 1.47 m / z

v_p = vo cos φ

v_p = 5.5 cos 15.5 = 5.30 m / s

cos θ = vₓ / v_p

sin θ = v_y / v_p

vₓ = v_p cos θ

v_y = v_p sin θ

vₓ = 5.30 cos 56.3 = 2.94 m / s

v_y = 5.30 sin 56.3 = 4.41 m / s

we already have the components of the initial velocity

v₀ = (2.94 i ^ + 4.41 j ^ + 1.47 k ^) m / s

let's find the acceleration on this axis (ax1) using Newton's second law

Fₓx = m aₓ₁

aₓ₁ = Fₓ / m

aₓ₁ = 220/100

aₓ₁ = 2.20 m / s²

Let's look for the velocity at the end of this interval (vx1)

Let's be careful if the initial velocity and they relate it has the same sense it must be added, but if the velocity and acceleration have the opposite direction it must be subtracted.

vₓ₁² = v₀ₓ² + 2 aₓ₁ (x₁-x₀)

let's calculate

vₓ₁² = 2.94² + 2 2.20 (14-8)

vₓ₁ = √35.04

vₓ₁ = 5.92 m / s

to the second interval

they relate it to xf

aₓ₂ = Fₓ₂ / m

aₓ₂ = 100/100

aₓ₂ = 1 m / s²

final speed

v_{xf}² = vₓ₁² + 2 aₓ₂ (x_f- x₁)

v_{xf}² = 5.92² + 2 1 (17-14)

v_{xf} =√41.05

v_{xf} = 6.41 m / s

We carry out the same calculation for each of the other axes.

Axis y

acceleration (a_{y1})

a_{y1} = F_y / m

a_{y1} = 460/100

a_[y1} = 4.60 m / s²

the velocity at the end of the interval (v_{y1})

v_{y1}² = v_{oy}² + 2 a_{y1{ (y₁ -y₀)

v_{y1}2 = 4.41² + 2 4.60 (-21 + 30)

v_{y1} = √102.25

v_{y1} = 10.11 m / s

second interval

acceleration (a_{y2})

a_{y2} = F_{y2} / m

a_{y2} = 260/100

a_{y2} = 2.60 m / s2

final speed

v_{yf}² = v_{y1}² + 2 a_{y2} (y₂ -y₁)

v_{yf}² = 10.11² + 2 2.60 (-27 + 21)

v_{yf} = √ 71.01

v_{yf} = 8.43 m / s

here there is an inconsistency in the problem if the body is at y₁ = -27m and passes the position y_f = -21 m with the relationship it must be contrary to the velocity

z axis

first interval, relate (a_{z1})

a_{z1} = F_{z1} / m

a_{z1} = -200/100

a_{z1} = -2 m / s

the negative sign indicates that the acceleration is the negative direction of the z axis

the speed at the end of the interval

v_{z1}² = v_{zo)² + 2 a_{z1} (z₁-z₀)

v_{z1}² = 1.47² + 2 (-2) (-7 + 4)

v_{z1} = √14.16

v_{z1} = -3.76 m / s

second interval, acceleration (a_{z2})

a_{z2} = F_{z2} / m

a_{z2} = 210/100

a_{z2} = 2.10 m / s2

final speed

v_{fz}² = v_{z1}² - 2 a_{z2} | z_f-z₁)

v_{fz}² = 3.14² - 2 2.10 (-3 + 7)

v_{fz} = √6.94

v_{fz} = 2.63 m / s

speed is v = ( 6.41 i^ + 8.43 j^ + 2.63 k^ ) m/s

5 0

3 years ago