Which has the same relationship as circle:cylinder? Triangle:sphere, square:window, rectangle:box, pentagon:counter

You hold a bucket in one hand. In the bucket is a 500 g rock. You swing the bucket so the rock moves in a vertical circle 2.2 m

slavikrds [6]

Answer:v=3.28 m/s

Explanation:

Given

mass of rock $m=500 gm$

diameter of circle $d=2.2 m$

radius $r=\frac{2.2}{2}=1.1 m$

At highest Point

$mg+N=\frac{mv^2}{r}$

At highest Point N=0 because mass is just balanced by centripetal Force

thus $mg=\frac{mv^2}{r}$

$v=\sqrt{gr}$

$v=\sqrt{9.8\times 1.1}$

$v=\sqrt{10.78}$

$v=3.28 m/s$

6 0

3 years ago

A ball is kicked at an angle of 35° with the ground.a) What should be the initial velocity of the ball so that it hits a target

stiks02 [169]

Answer:

a.18.5 m/s

b.1.98 s

Explanation:

We are given that

$\theta=35^{\circ}$

a.Let $v_0$ be the initial velocity of the ball.

Distance,x=30 m

Height,h=1.8 m

$v_x=v_0cos\theta=v_0cos35$

$v_y=v_0sin\theta=v_0sin35$

$x=v_0cos\theta\times t=v_0cos35\times t$

$t=\frac{30}{v_0cos35}$

$h=v_yt-\frac{1}{2}gt^2$

Substitute the values

$1.8=v_0sin35\frac{30}{v_0cos35}-\frac{1}{2}(9.8)(\frac{30}{v_0cso35})^2$

$1.8=30tan35-\frac{6574.6}{v^2_0}$

$\frac{6574.6}{v^2_0}=21-1.8=19.2$

$v^2_0=\frac{6574.6}{19.2}$

$v_0=\sqrt{\frac{6574.6}{19.2}}=18.5 m/s$

Initial velocity of the ball=18.5 m/s

b.Substitute the value then we get

$t=\frac{30}{18.5cos35}$

t=1.98 s

Hence, the time for the ball to reach the target=1.98 s

7 0

3 years ago

If ball 4 has a mass of 2 kg and it is 5m high, what will be its gravitational potential energy? (g=10 N/kg) *

Nastasia [14]

Explanation:

Gravitational potential energy

= mgh

= (2kg)(10N/kg)(5m)

= 100J.

3 0

3 years ago

Read 2 more answers

Which statement best describes wavelength?

xxMikexx [17]

Answer:

B

Explanation:

Wave length is the height perpendicular verically of a wave

5 0

2 years ago

Read 2 more answers

In 1994, Leroy Burrell of the United States set what was then a new world record for the men’s 100 m run. He ran the 1.00  102

vovikov84 [41]

Answer:

61.33 Kg

Explanation:

From the question given above, the following data were obtained:

Distance = 1×10² m

Time = 9.5 s

Kinetic energy (KE) = 3.40×10³ J

Mass (m) =?

Next, we shall determine the velocity Leroy Burrell. This can be obtained as follow:

Distance = 1×10² m

Time = 9.5 s

Velocity =?

Velocity = Distance / time

Velocity = 1×10² / 9.5

Velocity = 10.53 m/s

Finally, we shall determine the mass of Leroy Burrell. This can be obtained as follow:

Kinetic energy (KE) = 3.40×10³ J

Velocity (v) = 10.53 m/s

Mass (m) =?

KE = ½mv²

3.40×10³ = ½ × m × 10.53²

3.40×10³ = ½ × m × 110.8809

3.40×10³ = m × 55.44045

Divide both side by 55.44045

m = 3.40×10³ / 55.44045

m = 61.33 Kg

Thus, the mass of Leroy Burrell is 61.33 Kg

5 0

2 years ago