Levers pivot on the<br> A. fulcrum<br> B. resistance arm<br> C. effort arm<br> D.moon

a bus starting from rest moves with a uniform acceleration of 0.1 metre per second square for 2 minutes find the speed acquired

blondinia [14]

S ?
U 0m/s
V ?
A 0.1m/s^2
T 2min (120 sec)

S=ut+0.5at^2
S=0(120 sec)+0.5(0.1m/s^2)(120 sec)^2
S=720m

Distance double 720m*2=1440m

V^2=u^2+2as
V^2=(0)^2+2(0.1 m/s^2)(1440m)
V^2=288
V= square root of 288=12 root 2=16.97 to 2 decimal places

6 0

2 years ago

find a magnitude of the force such that if the act at right angle there resultant is √10N but if the act of 50° the resultant is

Readme [11.4K]

Explanation:

Let magnitude of the two forces be x and y.

Resultant at right angle R1= √15N) and at

60 degrees be R2= √18N.

Now, R1 = √(x² + y²) = √15,

R2= √(x² + y² +2xycos50) = √18.

So x² + y² = 15,

and x² + y² + 1.29xy = 18,

therefore 1.29xy = 3,

y = 3/1.29x.

y = 2.33/x

Now, x2 + (2.33/x)2 = 15,

x² + 5.45/x² = 15

multiply through by x²

x⁴ + 5.45 = 15x²

x⁴ - 15x2 + 5.45 = 0

Now find the roots of the equation, and later y. The two values of x will correspond to the

magnitudes of the two vectors.

Good luck

7 0

3 years ago

A rocket on Earth experiences an upward applied force from its thrusters. As a result of this force, the rocket accelerates upwa

gayaneshka [121]

Answer:

F=m(11.8m/s²)

For example, if m=10,000kg, F=118,000N.

Explanation:

There are only two vertical forces acting on the rocket: the force applied from its thrusters F, and its weight mg. So, we can write the equation of motion of the rocket as:

$F-mg=ma$

Solving for the force F, we obtain that:

$F=ma+mg=m(a+g)$

Since we know the values for a (2m/s²) and g (9.8m/s²), we have that:

$F= m(2m/s^{2}+9.8m/s^{2})\\\\F=m(11.8m/s^{2})$

From this relationship, we can calculate some possible values for F and m. For example, if m=10,000kg, we can obtain F:

$F=(10,000kg)(11.8m/s^{2})\\\\F=118,000N$

In this case, the force from the rocket's thrusters is equal to 118,000N.

5 0

3 years ago

An airplane has an effective wing surface area of 19.4 m2 that is generating the lift force. In level flight the air speed over

alexgriva [62]

Answer:

W =23807.68 N

Explanation:

given,

surface area of wing = 19.4 m²

speed over top wing = 67 m/s

speed under wing = 51 m/s

density of air = 1.3 kg/m³

weight of plane

From Bernoulli's principle

$P_1 + \dfrac{1}{2}\rho v_1^2 = P_2 + \dfrac{1}{2}\rho_2^2$

where 1 and 2 are two different locations at the same geo potential level

so if we call 1 the lower surface and 2 the upper surface,

we find the pressure differential, P₁ -P₂

$\Delta P =\dfrac{1}{2}\rho (v_2^2-v_1^2)$

$\Delta P =\dfrac{1}{2}\times 1.3 \times (67^2-51^2)$

$\Delta P =1227.2\ N/m^2$

then the force acting on the plane is

F=P A

F=1227.2 x 19.4

F =23807.68 N

weight of the plane

W =23807.68 N

7 0

2 years ago

Two brothers, Frank and Dion, are on roller skates and standing still when they push off from one another. Frank's mass is 103.4

tigry1 [53]

Around -3.5 ( might not be completely right ) Frank has a greater mass to he has a low velocity at first, Dion is around half of his weight so he doesn’t have to use as much force to get more than Frank

3 0

2 years ago

Levers pivot on the A. fulcrum B. resistance arm C. effort arm D.moon