Which situation will change the direction of the bicycle?

A wild pig attacks a hunter with a velocity of 5 m/s. At the instant when the pig is at a distance of 100 m the hunter shoots an

castortr0y [4]

Answer:

31.55 m/s

Explanation:

Let the initial velocity of the arrow is u metre per second.

Angle of projection, θ = 40 degree

range = 100 m

Use the formula for the range.

$R = \frac{u^{2}Sin2\theta }{g}$

100 = u^2 Sin(2 x 40) / 9.8

100 x 9.8 = u^2 Sin 80

u^2 = 995.11

u = 31.55 m/s

4 0

3 years ago

At the bottom of its path, the ball strikes a 2.30 kg steel block initially at rest on a frictionless surface. The collision is

Triss [41]

Answer:

(a). The speed of the ball after collision is 2.01 m/s.

(b). The speed of the block after collision 1.11 m/s.

Explanation:

Suppose, A steel ball of mass 0.500 kg is fastened to a cord that is 50.0 cm long and fixed at the far end. The ball is then released when the cord is horizontal.

Given that,

Mass of steel block = 2.30 kg

Mass of ball = 0.500 kg

Length of cord = 50.0 cm

We need to calculate the initial speed of the ball

Using conservation of energy

$\dfrac{1}{2}mv^2=mgl$

$v=\sqrt{2gl}$

Put the value into the formula

$u=\sqrt{2\times9.8\times50.0\times10^{-2}}$

$u=3.13\ m/s$

The initial speed of the ball $u_{1}=3.13\ m/s$

The initial speed of the block $u_{2}=0$

(a). We need to calculate the speed of the ball after collision

Using formula of collision

$v_{1}=(\dfrac{m_{1}-m_{2}}{m_{1}+m_{2}})u_{1}+(\dfrac{2m_{2}}{m_{1}+m_{2}})u_{2}$

Put the value into the formula

$v_{1}=(\dfrac{0.5-2.30}{0.5+2.30})\times3.13$

$v_{1}=-2.01\ m/s$

Negative sign shows the opposite direction of initial direction.

(b). We need to calculate the speed of the block after collision

Using formula of collision

$v_{2}=(\dfrac{2m_{1}}{m_{1}+m_{2}})u_{1}+(\dfrac{m_{1}-m_{2}}{m_{1}+m_{2}})u_{2}$

Put the value into the formula

$v_{2}=(\dfrac{2\times0.5}{0.5+2.30})\times3.13+0$

$v_{2}=1.11\ m/s$

Hence, (a). The speed of the ball after collision is 2.01 m/s.

(b). The speed of the block after collision 1.11 m/s.

8 0

3 years ago

You're holding an open house for Brenda. Walt and Mary walk in and ask you for a flyer that gives details about the property. At

lapo4ka [179]

Walt and Mary are my Customers at this point.

To construct proper client relationships you need to: greet clients and approach them in a way that is herbal and suits the character scenario. show customers that you recognize what their desires are. be given that a few humans may not want your merchandise and concentrate on constructing relationships with people who do.

Use the time period clients, with an apostrophe before the “s” to expose a possessive form for a single consumer. Use the time period clients', with an apostrophe after the “s” to show the possessive plural shape of a couple of patrons. Do now not use an apostrophe if there's no possessive indication needed.

The definition of a consumer is a person who buys services or products from a store, restaurant, or different retail vendor. An example of a patron is someone who is going to an electronics save and buys a tv. (Casual) a person, mainly one engaging in a few types of interaction with others.

Learn more about the Customers herehttps://brainly.com/question/24448358

#SPJ1

8 0

1 year ago

Two particles in a high-energy accelerator experiment are approaching each other head-on, each with a speed of 0.9520c as measur

Over [174]

Answer:

the magnitude of the velocity of one particle relative to the other is 0.9988c

Explanation:

Given the data in the question;

Velocities of the two particles = 0.9520c

Using Lorentz transformation

Let relative velocity be W, so

v $_r$ = ( u + v ) / ( 1 + ( uv / c²) )

since each particle travels with the same speed,

u = v

so

v $_r$ = ( u + u ) / ( 1 + ( u×u / c²) )

v $_r$ = 2(0.9520c) / ( 1 + ( 0.9520c )² / c²) )

we substitute

v $_r$ = 1.904c / ( 1 + ( (0.906304 × c² ) / c²) )

v $_r$ = 1.904c / ( 1 + 0.906304 )

v $_r$ = 1.904c / 1.906304

v $_r$ = 0.9988c

Therefore, the magnitude of the velocity of one particle relative to the other is 0.9988c

5 0

3 years ago

HELP!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!pls

barxatty [35]

I think this is the solution:

1: U-1, F,-4
2: Na-6, Mo-1, O-4
3: Bi-1, O-1, C-1, I-1
4: In-9, N-1
5: N-2, H-4, S-1, C-1
6: Ge- 15, N-4
7: N-1, H-4, C-1, I-1, O-3
8: H-7, F-1
9: N-1, O-5, H-1, S-1
10: H-8
11: Nb-1, O-1, C-1, I-3
12: C-3, F-3, S-1, O-3, H-1
13: Ag-1, C-1, N-1, O-1
14: Pb-6, H-1, As-1, O-4

5 0

2 years ago