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Leona [35]
2 years ago
13

Describe how water can be both physical and chemical weathering.

Physics
1 answer:
Lynna [10]2 years ago
5 0

Explanation:

Both because water can fall in holes and then freeze. However, water can also be chemically react with other elements and substances to wear something away.

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Two equal forces are applied to a door. The first force is applied at the midpoint of the door, the second force is applied at t
Orlov [11]

Answer:

D) the second at the doorknob

Explanation:

The torque exerted by a force is given by:

\tau = Fdsin \theta

where

F is the magnitude of the force

d is the distance between the point of application of the force and the centre of rotation

\theta is the angle between the direction of the force and d

In this problem, we have:

- Two forces of equal magnitude F

- Both forces are perpendicular to the door, so \theta=90^{\circ}, sin \theta=1

- The first force is exerted at the midpoint of the door, while the 2nd force is applied at the doorknob. This means that d is the larger for the 2nd force

--> therefore, the 2nd force exerts a greater torque

4 0
3 years ago
Based on observations, the speed of a jogger can be approximated by the relation v 5 7.5(1 2 0.04x) 0.3, where v and x are expre
castortr0y [4]

Answer:

solution:

to find the speed of a jogger use the following relation:  

V

=

d

x

/d

t

=

7.5

×m

i

/

h

r

...........................(

1

)  

in Above equation in x and t. Separating the variables and integrating,

∫

d

x

/7.5

×=

∫

d

t

+

C

or

−

4.7619  

=

t

+

C

Here C =constant of integration.   

x

=

0  at  t

=

0

, we get:  C

=

−

4.7619

now we have the relation to find the position and time for the jogger as:

−

4.7619  =

t

−

4.7619

.

.

.

.

.

.

.

.

.

(

2

)

Here

x  is measured in miles and  t  in hours.

(a) To find the distance the jogger has run in 1 hr, we set t=1 in equation (2),    

     to get:

      = −

4.7619  

      =  

1

−

4.7619

      = −

3.7619

  or  x

=

7.15

m

i

l

e

s

(b) To find the jogger's acceleration in   m

i

l

/

differentiate  

     equation (1) with respect to time.

     we have to eliminate x from the equation (1) using equation (2).  

     Eliminating x we get:

     v

=

7.5×

     Now differentiating above equation w.r.t time we get:

      a

=

d

v/

d

t

       =

−

0.675

/

      At  

      t

=

0

      the joggers acceleration is :

       a

=

−

0.675

m

i

l

/

        =

−

4.34

×

f

t

/  

(c)  required time for the jogger to run 6 miles is obtained by setting  

        x

=

6  in equation (2).  We get:

        −

4.7619

(

1

−

(

0.04

×

6  )

)^

7

/

10=

t

−

4.7619

         or

         t

=

0.832

h

r

s

6 0
3 years ago
a missile is moving 1810 m/s at a 20.0 degree angle. it needs to hit a target 19,500 m away in a 32.0 degree direction in 9.20 s
Ket [755]

Answer:

112 m/s², 79.1°

Explanation:

In the x direction, given:

x₀ = 0 m

x = 19,500 cos 32.0° m

v₀ = 1810 cos 20.0° m/s

t = 9.20 s

Find: a

x = x₀ + v₀ t + ½ at²

19,500 cos 32.0° = 0 + (1810 cos 20.0°) (9.20) + ½ a (9.20)²

a = 21.01 m/s²

In the y direction, given:

y₀ = 0 m

y = 19,500 sin 32.0° m

v₀ = 1810 sin 20.0° m/s

t = 9.20 s

Find: a

y = y₀ + v₀ t + ½ at²

19,500 sin 32.0° = 0 + (1810 sin 20.0°) (9.20) + ½ a (9.20)²

a = 109.6 m/s²

The magnitude of the acceleration is:

a² = ax² + ay²

a² = (21.01)² + (109.6)²

a = 112 m/s²

And the direction is:

θ = atan(ay / ax)

θ = atan(109.6 / 21.01)

θ = 79.1°

5 0
3 years ago
A small plastic ball with a mass of 7.00 10-3 kg and with a charge of +0.155 µC is suspended from an insulating thread and hangs
adell [148]

Answer:

the magnitude of the charge Q on each plate is 3.053 *10^{-8} \ C

Explanation:

Given that :

mass (m) = 7.00 *10 ^{-3} \ kg

charge (q) = +0.155 µC = +0.155 *10^{-6}\ C

angle \theta = 30^0 \ C

Area A on each plate = 0.0135 m²

From the diagram below;

tan \ \theta = \frac{Eq}{mg}    ----- equation (1)

Also by using Gauss Law ;

Q = \epsilon_0 \phi

Q = \epsilon_0EA     ----- equation (2)

Combination equation 1 and 2 together ; we have

Q = \frac{\epsilon_0\ * \ m\ *\ g \ \ * \ tan \theta \ * \ A}{q}

Q = \frac{(8.85*10^{-12}C^2/N.m^2 )\ * \ (7.00*10^{-3} kg)\ *\ (9.8 m/s^2) \ \ * \ tan \(30 \ * \ (0.0135 m^2)}{0.155*10^{-6}\ C}

Q = 3.053 *10^{-8} \ C

8 0
3 years ago
Question 2 (Multiple Choice Worth 3 points)
zmey [24]
Surveys are considered the most reliable way to gather data
7 0
3 years ago
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