What phenotypes would you expect from a cross between a red bull and a white cow?

A 10- kilogram block is pushed across a horizontal surface with a horizontal force of 20 N against a friction force of 10 N. The

Temka [501]

Answer:

$1m/s^2$

Explanation:

Mass of block=10 kg

Applied horizontal force =F=20 N

Friction force=f=10 N

We have to find the acceleration of block.

Net force=Applied horizontal force-friction force

$ma=F-f$

Where F= Horizontal force

f=Friction force

m=Mass of object

a=Acceleration of object

$10a=20-10=10$

$a=\frac{10}{10}=1 m/s^2$

Hence, the acceleration of the block= $1m/s^2$

4 0

3 years ago

The vector sum of the forces acting on the beam is zero, and the sum of the moments about the left end of the beam is zero. (a)

Marysya12 [62]

This question is incomplete, the complete question is;

The vector sum of the forces acting on the beam is zero, and the sum of the moments about the left end of the beam is zero.

(a) Determine the forces and and the couple

(b) Determine the sum of the moments about the right end of the beam.

(c) If you represent the 600-N force, the 200-N force, and the 30 N-m couple by a force F acting at the left end of the beam and a couple M, what is F and M?

Answer:

a)

the x-component of the force at A is $A_{x}$ = 0

the y-component of the force at A is $A_{y}$ = 400 N

the couple acting at A is; $M_{A}$ = 146 N-m

b)

the sum of the momentum about the right end of the beam is; ∑ $M_{R}$ = 0

c)

the equivalent force acting at the left end is; F = -400J ( N)

the couple acting at the left end is; M = - 146 N-m

Explanation:

Given that;

The sum of the forces acting on the beam is zero ∑f = 0

Sum of the moments about the left end of the beam is also zero ∑ $M_{L}$ = 0

Vector force acting at A, $F_{A}$ = $A_{x}i$ + $A_{y}j$

Now, From the image, we have;

a)

∑f = 0

$F_{A}$ - 600j + 200j = 0i + 0j

$A_{x}i$ + $A_{y}j$ - 600j + 200j = 0i + 0j

$A_{x}i$ + ( $A_{y}$ - 400)j = 0i + 0j

now by equating i- coefficients'

$A_{x}$ = 0

so, the x-component of the force at A is $A_{x}$ = 0

also by equating j-coefficient

$A_{y}$ - 400 = 0

$A_{y}$ = 400 N

hence, the y-component of the force at A is $A_{y}$ = 400 N

we also have;

∑ $M_{L}$ = 0

$M_{A}$ - ( 30 N-m ) - ( 0.380 m )( 600 N ) + ( 0.560 m )( 200 N ) = 0

$M_{A}$ - 30 N-m - 228 N-m + 112 Nm = 0

$M_{A}$ - 146 N-m = 0

$M_{A}$ = 146 N-m

Therefore, the couple acting at A is; $M_{A}$ = 146 N-m

b)

The sum of the moments about right end of the beam is;

∑ $M_{R}$ = (0.180 m)(600N) - (30 N-m) - ( 0.56 m)( $A_{y}$ ) + $M_{A}$

∑ $M_{R}$ = (108 N-m) - (30 N-m) - ( 0.56 m)(400 N ) + 146 N-m

∑ $M_{R}$ = (108 N-m) - (30 N-m) - ( 224 N-m ) + 146 N-m

∑ $M_{R}$ = 0

Therefore, the sum of the momentum about the right end of the beam is; ∑ $M_{R}$ = 0

c)

The 600-N force, the 200-N force and the 30 N-m couple by a force F which is acting at the left end of the beam and a couple M.

The equivalent force at the left end will be;

F = -600j + 200j (N)

F = -400J ( N)

Therefore, the equivalent force acting at the left end is; F = -400J ( N)

Also couple acting at the left end

M = -(30 N-m) + (0.560 m)( 200N) - ( 0.380 m)( 600 N)

M = -(30 N-m) + (112 N-m) - ( 228 N-m))

M = 112 N-m - 258 N-m

M = - 146 N-m

Therefore, the couple acting at the left end is; M = - 146 N-m

7 0

3 years ago

A force of 6.0 N gives a 2.0 kg block an acceleration of 3.0

Semenov [28]

Explanation:

The Net Force of the object can be written by:

Fnet = ma

where m is the mass of the object in kg

a is the acceleration of the object in m/s^2

Hence by applying the formula we get:

Fnet = (2.0)(3.0)

= 6N

We also know that Net force is also the sum of all forces acting on an object. In this case Friction and the Pushing Force is acting on the object. Hence we can write that:

Fnet = Pushing Force + (-Friction)

6N = 6N - Friction

Friction = 0N

Hence the force of friction is 0N.

7 0

3 years ago

a car with a mass of 100 kg is stopped on the side of the road after getting a flat tire. the two people that were riding in the

notsponge [240]

Answer:

Please mark me brainliest and thank me and rate me

3 0

3 years ago

A river flows due east at 1.60 m/s. A boat crosses the river from the south shore to the north shore by maintaining a constant v

Citrus2011 [14]

Answer:

part (a) $v\ =\ 10.42\ at\ 81.17^o$ towards north east direction.

part (b) s = 46.60 m

Explanation:

Given,

velocity of the river due to east = $v_r\ =\ 1.60\ m/s.$
velocity of the boat due to the north = $v_b\ =\ 10.3\ m/s.$

part (a)

River is flowing due to east and the boat is moving in the north, therefore both the velocities are perpendicular to each other and,

Hence the resultant velocity i,e, the velocity of the boat relative to the shore is in the North east direction. velocities are the vector quantities, Hence the resultant velocity is the vector addition of these two velocities and the angle between both the velocities are $90^o$

Let 'v' be the velocity of the boat relative to the shore.

$\therefore v\ =\ \sqrt{v_r^2\ +\ v_b^2}\\\Rightarrow v\ =\ \sqrt{1.60^2\ +\ 10.3^2}\\\Rightarrow v\ =\ 10.42\ m/s.$

Let $\theta$ be the angle of the velocity of the boat relative to the shore with the horizontal axis.

Direction of the velocity of the boat relative to the shore. $\therefore Tan\theta\ =\ \dfrac{v_b}{v_r}\\\Rightarrow Tan\theta\ =\ \dfrac{10.3}{1.60}\\\Rightarrow \theta\ =\ Tan^{-1}\left (\dfrac{10.3}{1.60}\ \right )\\\Rightarrow \theta\ =\ 81.17^o$

part (b)

Width of the shore = w = 300m

total distance traveled in the north direction by the boat is equal to the product of the velocity of the boat in north direction and total time taken

Let 't' be the total time taken by the boat to cross the width of the river. $\therefore w\ =\ v_bt\\\Rightarrow t\ =\ \dfrac{w}{v_b}\\\Rightarrow t\ =\ \dfrac{300}{10.3}\\\Rightarrow t\ =\ 29.12 s$

Therefore the total distance traveled in the direction of downstream by the boat is equal to the product of the total time taken and the velocity of the river $\therefore s\ =\ u_rt\\\Rightarrow s\ =\ 1.60\times 29.12\\\Rightarrow s\ =\ 46.60\ m$

7 0

3 years ago