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fiasKO [112]
3 years ago
9

A push on a 1-kilogram brick accelerates the brick. Neglecting friction, to equally accelerate a 10-kilogram brick, one would ha

ve to push
A. with 100 times as much force.
B. with 10 times as much force.
C. with just as much force.
D. with the amount of force.
E. none of the above
Physics
1 answer:
olga55 [171]3 years ago
4 0
I believe the answer is C. With just as much force.

Hope this helps
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The distance between the centers of the wheels of a motorcycle is 156 cm. The center of mass of the motorcycle, including the ri
Drupady [299]

Answer:

a = 9.86 m/s²

Explanation:

given,

distance between the centers of wheel = 156 cm

center of mass of motorcycle including rider = 77.5 cm

horizontal acceleration of motor cycle = ?

now,

The moment created by the wheels must equal the moment created by gravity.

take moment about wheel as it touches the ground, here we will take horizontal distance between them.

then, take the moment around the center of mass. Since the force on the ground from the wheels is horizontal, we need the vertical distance.

now equating both the moment

m g d = F h

d is the horizontal distance

h is the vertical distance

m g d = m a h

term of mass get eliminated

g d = a h

so,

a = \dfrac{g\ d}{h}

a = \dfrac{9.8\times 0.78}{0.775}

a = 9.86 m/s²

4 0
3 years ago
Particle A of charge 3.06 10-4 C is at the origin, particle B of charge -5.70 10-4 C is at (4.00 m, 0), and particle C of charge
Alex

Answer:

F_net = 26.512 N

Explanation:

Given:

Q_a = 3.06 * 10^(-4 ) C

Q_b = -5.7 * 10^(-4 ) C

Q_c = 1.08 * 10^(-4 ) C

R_ac = 3 m

R_bc = sqrt (3^2 + 4^2) = 5m

k = 8.99 * 10^9

Coulomb's Law:

F_i = k * Q_i * Q_j / R_ij^2

Compute F_ac and F_bc :

F_ac = k * Q_a * Q_c / R^2_ac

F_ac =  8.99 * 10^9* ( 3.06 * 10^(-4 ))* (1.08 * 10^(-4 )) / 3^2

F_ac = 33.01128 N

F_bc = k * Q_b * Q_c / R^2_bc

F_bc =  8.99 * 10^9* ( 5.7 * 10^(-4 ))* (1.08 * 10^(-4 )) / 5^2  

F_bc = - 22.137 N

Angle a is subtended between F_bc and y axis @ C

cos(a) = 3 / 5

sin (a) = 4 / 5

Compute F_net:

F_net = sqrt (F_x ^2 + F_y ^2)

F_x = sum of forces in x direction:

F_x = F_bc*sin(a) = 22.137*(4/5) = 17.71 N

F_y = sum of forces in y direction:

F_y = - F_bc*cos(a) + F_ac = - 22.137*(3/5) + 33.01128 = 19.72908 N

F_net = sqrt (17.71 ^2 + 19.72908 ^2) = 26.5119 N

Answer: F_net = 26.512 N

5 0
4 years ago
55. (III) A uniform circular plate of radius 2R has a circular hole of radius R cut out of it. The center of the smaller circle
Aleksandr [31]

Answer:

P = 0.27R from the center

Explanation:

Given,

The radius of the uniform circular plate, R = 2R

The radius of the hole, r = R

The center of the smaller circle from the center is, d = 0.8R

The center of mass of a circular disc with a hole in it given by the formula

                               P = dr²/R² - r²

Where P is the distance from the center of mass located in the line joining the two centers opposite to the hole.

Substituting the given values in the above equation,

                               P = 0.8R x R² / 4R² - R²

                                  = 0.27R³/R²

                                  = 0.27R

Hence the center of mass of plate is at a distant  P = 0.27R from the center

6 0
3 years ago
9. How does the length of the hypotenuse in a right triangle relate to the lengths of the legs? (2 points)
konstantin123 [22]
<span>The pythagorean theorem addresses the length of the hypotenuse in relation to the length of the legs. The square root of the length of the hypotenuse is equal to the sum of one leg squared plus the other leg squared. In other words, A squared plus B squared equals C squared where A and B are the lengths of the legs of the triangle and C is the length of the hypotenuse.</span>
6 0
3 years ago
Read 2 more answers
What function does the shape of the star-nosed mole’s nose serve?
horsena [70]
Hello!

The shape of the star-nosed mole's nose helps it to take away the excess heat.

Letter b)

Hugs!
6 0
3 years ago
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