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2 years ago

Suppose you give a 10 Newton push to Ryan on skis (he weighs 50 kg), how much will he accelerate?

1 answer:

3 0

Well we can just use F=ma. The force is 10N, the mass is 50 kg, solve for a. Well since we kg and N, no conversion is necessary. So just plugging in the numbers, we get

10N = 50 kg · a

$\frac{10N}{50kg}=a$

A newton is just $\frac{kg·m}{s^{2}}$

$a=\frac{\frac{10kg·m}{s^{2}}}{50kg}$

The s^2 and 50 kg you multiply

$a=\frac{10kg·m}{50kg·s^{2}}$

The kg's cancel and 10/50 is 1/5

$\frac{1}{5}·\frac{m}{s^{2}}$

So the acceleration is 1/5 m/s^2

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On the way to school, the bus speeds up from 20 m/s to 36 m/s in 4 seconds. What distance

Tresset [83]

Answer B. 112 m

Step-by-Step Explanation

initial velocity u = 20 m /s
final velocity v = 36 m /s
time taken t = 4 s
acceleration = (v - U) / t
= (36 - 20) / 4
a=4m/s2
from the formula
7-u2=2as , sis distance covered
putting the values
362-202=2×4×s
1296 - 400 = 8 x S
S= 112 m

4 0

2 years ago

A car travels north along a straight highway at an average speed of 85 km/h. After driving 2.0 km, the car passes a gas station

jenyasd209 [6]

Constant speed is the answer

5 0

3 years ago

A mass of 1 slug, when attached to a spring, stretches it 2 feet and then comes to rest in the equilibrium position. Starting at

Vesna [10]

Answer:

$Y=(\dfrac{3}{16}+t \dfrac{3}{8})e^{-2t}-\dfrac{3}{16}cos 4t$

Explanation:

Given that m= 1 slug and given that spring stretches by 2 feet so we can find the spring constant K

mg=k x

1 x 32= k x 2

K=16

And also give that damping force is 8 times the velocity so damping constant C=8.

We know that equation for spring mass system

my''+Cy'+Ky=F

Now by putting the values

1 y"+8 y'+ 16y=6 cos 4 t ----(1)

The general solution of equation Y=CF+IP

Lets assume that at steady state the equation of y will be

y(IP)=A cos 4t+ B sin 4t

To find the constant A and B we have to compare this equation with equation 1.

Now find y' and y" (by differentiate with respect to t)

y'= -4A sin 4t+4B cos 4t

y"=-16A cos 4t-16B sin 4t

Now put the values of y" , y' and y in equation 1

1 (-16A cos 4t-16B sin 4t)+8( -4A sin 4t+4B cos 4t)+16(A cos 4t+ B sin 4t)=6sin4 t

So by comparing the coefficient both sides

-16A+32B+16A=0 So B=0

-16 B-32 A+16B=6 So A=-3/16

y=-3/16 cos 4t

Now to find the CF of differential equation 1

y"+8 y'+ 16y=6 cos 4 t

Homogeneous version of above equation

$m^2+8m+16=0$

So $CF =(C_1+tC_2)e^{-2t}$

So the general equation

$Y=(C_1+tC_2)e^{-2t}-3/16 cos 4t$

Given that t=0 Y=0 So

$C_1=\dfrac{3}{16}$

t=0 Y'=0 So

$C_2 =\dfrac{3}{8}$

$Y=(\dfrac{3}{16}+t \dfrac{3}{8})e^{-2t}-\dfrac{3}{16}cos 4t$

The above equation is the general equation for motion.

3 0

3 years ago

Read each description below. Choose the force diagram (free-body diagram) that best represents the description. You may neglect

Papessa [141]

A. Diagram A
B. Diagram C & D
C. Diagram B
D. Diagram C & D
E. Diagram B
F. Diagram C & D
These are simplified representations of an object's body and the force vectors acting on it. Some of the main forces that are involve are normal force, friction, push or pull and gravity.

5 0

3 years ago

A 16.0kg canoe moving to the left at 12.5m/s makes an elastic head-on collision with a 14.0kg raft moving to the right at 16.0m/

kherson [118]

The canoe is moving at 14.1 m/s to the right after the collision.

Explanation:

According to the law of conservation of momentum, in absence of external forces the total momentum of the system must be conserved before and after the collision. So we can write:

$p_i = p_f\\m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2$

where:

$m_1 = 16.0 kg$ is the mass of the canoe

$u_1 = -12.5 m/s$ is the initial velocity of canoe (we take right as positive direction, and since the canoe is moving to the left, its velocity is negative)