OK draw a diagram - you have the force from (b) acting sown the slope and a component of the weight. Use F=ma to get the deceleration and then use SUVAT. Post your working if this doesn't work.

friction = 900N

braking = 3100N

total = 4000N

w= mxg = 1000*9.81= 9810N

total = 9810 + 4000 = 13810N

force/mass = 13810/1000 = 13.81ms^-2

then using v^2 = u^2 + 2as, i get s as 14 but it is incorrect

Explanation:

4 0

3 years ago

In a science fiction novel two enemies, Bonzo and Ender, are fighting in outer spce. From stationary positions, they push agains

lara [203]

Answer:

ratio of the masses of the two enemies = 1.7

Explanation:

Applying the law of conservation of momentum,

Momentum fo Bonzo = Momentum of Ender

mv = m'v'................. Equation 1

Where m = mass of Bonzo, v = velocity of Bonzo, m' = mass of Ender, v' = velocity of Ender

m/m' = v'/v............... Equation 2

Where m/m' = ratio of the masses of the two enemies

Given: v = 2.0 m/s, v' = 3.4 m/s

Substitute into equation 2

m/m' = 3.4/2

m/m' = 1.7

Hnece, ratio of the masses of the two enemies = 1.7

7 0

3 years ago

A vaulter holds a 26.90-N pole in equilibrium by exerting an upward force U with her leading hand and a downward force D with he

ivolga24 [154]

Answer:

Check attachment for better understanding

Explanation:

Given that

a= 0.75m

b=1.31m

c= 2.2m

Weight of pole is 26.90

Then, Fg = Weight = 26.90

Using Equilibrium of forces

ΣFy = 0

U — D — Fg = 0

U — D = Fg

U — D = 26.9

To calculate U,

We will take moment about point A.

ΣMa = 0

Let the clockwise moment be positive and anti-clockwise be negative

Fg(a+b) — U(a) = 0

26.9(1.31+0.75) —0.75U = 0

26.9(2.06) = 0.75U

0.75U = 55.414

U = 55.414/0.75

U = 73.89 N

To calculate D,

U — D = 26.9

73.89—D =26.9

73.89—26.9 = D

D = 46.99N

3 0

3 years ago

Read 2 more answers

In a certain clock, a pendulum of length L1 has a period T1 = 0.95s. The length of the pendulum

gulaghasi [49]

Answer:

Ratio of length will be $\frac{L_2}{L_1}=1.108$

Explanation:

We have given time period of the pendulum when length is $L_1$ is $T_1=0.95sec$

And when length is $L_2$ time period $T_2=1sec$

We know that time period is given by

$T=2\pi \sqrt{\frac{L}{g}}$

So $0.95=2\pi \sqrt{\frac{L_1}{g}}$ ----eqn 1

And $1=2\pi \sqrt{\frac{L_2}{g}}$ -------eqn 2

Dividing eqn 2 by eqn 1

$\frac{1}{0.95}=\sqrt{\frac{L_2}{L_1}}$

Squaring both side

$\frac{L_2}{L_1}=1.108$

8 0

4 years ago