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abruzzese [7]
3 years ago
8

Explain how attaching the key to a piece of wood could prevent the key from sinking

Physics
2 answers:
amm18123 years ago
5 0
Wood floats, whereas the key doesn’t. If you attach the 2 together than the wood would prevent the key from sinking.
Free_Kalibri [48]3 years ago
3 0

It could prevent the key from sinking because the wood would float.

It floats because it weighs less than amount of water it would have to push out of the glass if it sank. Wood, cork, and ice are all less dense than water, and they float; rocks are more dense, so they sink. A key would also be more dense causing it to sink.

Hope this helps,

Davinia.

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If a box is pulled with a force of 100 N at an angle of 25
love history [14]

The X and Y components of the force are 90.63 Newton and 42.26 Newton respectively.

<u>Given the following data:</u>

  • Force = 100 Newton.
  • Angle of inclination = 25°

To determine the X and Y components of the force:

<h3>The horizontal component (X) of a force:</h3>

Mathematically, the horizontal component of a force is given by this formula:

F_x = Fcos \theta\\\\F_x =100 \times cos25\\\\F_x =100 \times 0.9063

Fx = 90.63 Newton.

<h3>The vertical component (Y) of tensional force:</h3>

Mathematically, the vertical component of a force is given by this formula:

F_y = Fsin \theta\\\\F_y =100 \times sin25\\\\F_y =100 \times 0.4226

Fy = 42.26 Newton.

Read more on horizontal component here: brainly.com/question/4080400

6 0
2 years ago
Imagine two billiard balls on a pool table. Ball A has a mass of 7 kilograms and ball B has a mass of 2 kilograms. The initial v
wlad13 [49]
1) In a perfectly inelastic collision, the two balls stick together after the collision. In this type of collision, the total kinetic energy of the system is not conserved, while the total momentum is conserved.
If we callv_f the final velocity of the two balls that stick together, the conservation of the total momentum before and after the collision can be written as
m_a v_{Ai} + m_b v_{Bi} = (m_A+m_B)v_f (1)
where
m_A=7 kg is the mass of ball A
m_B=2 kg is the mass of ball B
v_{Ai}=6 m/s is the initial velocity of ball A
v_{Bi}=-12 m/s is the initial velocity of ball B (taken with a negative sign, since it goes in the opposite direction of ball A)

If we solve (1) to find v_f, we find that the final velocity of the balls is
v_f= \frac{m_Av_{Ai}+m_Bv_{Bi}}{m_A+m_B}= \frac{(7\cdot 6)+(2 \cdot (- 12))}{7+2}= \frac{18}{9}=2 m/s
and the positive sign means the two balls are going to the right.


2) I assume here we are talking about an elastic collision. In this case, both total momentum and total kinetic energy are conserved:
m_A v_{Ai}+m_B v_{Bi} = m_A v_{fA} + m_B v_{fB}
\frac{1}{2}m_A v_{Ai}^2+ \frac{1}{2}m_B v_{Bi}^2= \frac{1}{2}m_Av_{fA}^2+ \frac{1}{2}m_B v_{fB}^2
where
v_{fA} is the final velocity of ball A
v_{fB} is the final velocity of ball B

If we solve simultaneously the two equations, we find:
v_{fA}= \frac{v_{Ai}(m_A-m_B)+2m_Bv_{Bi}}{m_A+m_B} = \frac{(6)(7-2)+2(2)(-12)}{7+2}=-2 m/s
v_{fB}= \frac{v_{Bi}(m_B-m_A)+2m_Av_{Ai}}{m_A+m_B} = \frac{(-12)(2-7)+2(7)(6)}{7+2}= \frac{144}{9}=16 m/s
So, after the collision, ball A moves to the left with velocity v=-2 m/s and ball B moves to the right with velocity v=16 m/s.

3) The total momentum before and after the collision is conserved.
In fact, the total momentum before the collision is:
p_i = m_A v_{A} + m_B v_{fB} = (7\cdot 6)+(2 \cdot (-12))=42-24=18 m/s
and the total momentum after the collision is:
p_f = m_A v_{A} + m_B v_{fB} = (7\cdot (-2))+(2 \cdot 16)=-14+32=18 m/s

3 0
3 years ago
Which of the following best explains how earth is heated through radiation
lawyer [7]

Answer:

The sun touches earth during daytime and the suns rays heat our earth giving us heat. The sun heating the earth is also considered radiation.

Explanation:

6 0
3 years ago
Running at 1.55 m/s, Bruce, the 40.0 kg quarterback, collides with Biff, the 90.0 kg tackle, who is traveling at 7.0 m/s in the
Margarita [4]

Answer:Bruce is knocked backwards at  

14

m

s

.

Explanation:

This is a problem of momentum (

→

p

) conservation, where

→

p

=

m

→

v

and because momentum is always conserved, in a collision:

→

p

f

=

→

p

i

We are given that  

m

1

=

45

k

g

,  

v

1

=

2

m

s

,  

m

2

=

90

k

g

, and  

v

2

=

7

m

s

The momentum of Bruce (

m

1

) before the collision is given by

→

p

1

=

m

1

v

1

→

p

1

=

(

45

k

g

)

(

2

m

s

)

→

p

1

=

90

k

g

m

s

Similarly, the momentum of Biff (

m

2

) before the collision is given by

→

p

2

=

(

90

k

g

)

(

7

m

s

)

=

630

k

g

m

s

The total linear momentum before the collision is the sum of the momentums of each of the football players.

→

P

=

→

p

t

o

t

=

∑

→

p

→

P

i

=

→

p

1

+

→

p

2

→

P

i

=

90

k

g

m

s

+

630

k

g

m

s

=

720

k

g

m

s

Because momentum is conserved, we know that given a momentum of  

720

k

g

m

s

before the collision, the momentum after the collision will also be  

720

k

g

m

s

. We are given the final velocity of Biff (

v

2

=

1

m

s

) and asked to find the final velocity of Bruce.

→

P

f

=

→

p

1

f

+

→

p

2

f

→

P

f

=

m

1

v

1

f

+

m

2

v

2

f

Solve for  

v

1

:

v

1

f

=

→

P

f

−

m

2

v

2

f

m

1

Using our known values:

v

1

f

=

720

k

g

m

s

−

(

90

k

g

)

(

1

m

s

)

45

k

g

v

1

f

=

14

m

s

∴

Bruce is knocked backwards at  

14

m

s

.

Explanation:

5 0
3 years ago
What is the lowest temperature possible
dimulka [17.4K]

At the physically impossible-to-reach temperature of zero kelvin, or minus 459.67 degrees Fahrenheit (minus 273.15 degrees Celsius), atoms would stop moving. As such, nothing can be colder than absolute zero on the Kelvin scale.


3 0
3 years ago
Read 2 more answers
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