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Mariana [72]
2 years ago
15

A box was pushed across a floor a distance of 6.5m against a frictional resistance of 7.3N. what work was done​

Physics
1 answer:
Svet_ta [14]2 years ago
8 0

Answer:

I think the answer is friction

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In springboard diving, the diver strides out to the end of the board, takes a jump onto its end, and uses the resultant spring-l
Ksenya-84 [330]

Answer:

10.4 m/s

Explanation:

The problem can be solved by using the following SUVAT equation:

v=u+at

where

v is the final velocity

u is the initial velocity

a is the acceleration

t is the time

For the diver in the problem, we have:

u=+6.3 m/s is the initial velocity (positive because it is upward)

a=g=-9.8 m/s^2 is the acceleration of gravity (negative because it is downward)

By substituting t = 1.7 s, we find the velocity when the diver reaches the water:

v=+6.3 + (-9.8)(1.7)=-10.4 m/s

And the negative sign means that the direction is downward: so, the speed is 10.4 m/s.

3 0
3 years ago
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The force needed to keep a car from skidding on a curve varies inversely as the radius of the curve and jointly as the weight of
motikmotik

Explanation:

It is given that, the force needed to keep a car from skidding on a curve varies inversely as the radius of the curve and jointly as the weight of the car and the square of the car's speed such that,

F\propto \dfrac{mgv^2}{r}

F=\dfrac{kmgv^2}{r}

mg is the weight of the car

r is the radius of the curve

v is the speed of the car

Case 1.

F = 640 pounds

Weight of the car, W = mg = 2600 pound

Radius of the curve, r = 650 ft

Speed of the car, v = 40 mph

640=\dfrac{k(2600)(40)^2}{650}

k = 0.1

Case 2.

Radius of the curve, r = 750 ft

Speed of the car, v = 30 mph

F=\dfrac{0.1\times 2600\times (30)^2}{750}

F = 312 N

Hence, this is the required solution.

6 0
2 years ago
Zero, a hypothetical planet, has a mass of 5.3 x 1023 kg, a radius of 3.3 x 106 m, and no atmosphere. A 10 kg space probe is to
Andrej [43]

(a) 3.1\cdot 10^7 J

The total mechanical energy of the space probe must be constant, so we can write:

E_i = E_f\\K_i + U_i = K_f + U_f (1)

where

K_i is the kinetic energy at the surface, when the probe is launched

U_i is the gravitational potential energy at the surface

K_f is the final kinetic energy of the probe

U_i is the final gravitational potential energy

Here we have

K_i = 5.0 \cdot 10^7 J

at the surface, R=3.3\cdot 10^6 m (radius of the planet), M=5.3\cdot 10^{23}kg (mass of the planet) and m=10 kg (mass of the probe), so the initial gravitational potential energy is

U_i=-G\frac{mM}{R}=-(6.67\cdot 10^{-11})\frac{(10 kg)(5.3\cdot 10^{23}kg)}{3.3\cdot 10^6 m}=-1.07\cdot 10^8 J

At the final point, the distance of the probe from the centre of Zero is

r=4.0\cdot 10^6 m

so the final potential energy is

U_f=-G\frac{mM}{r}=-(6.67\cdot 10^{-11})\frac{(10 kg)(5.3\cdot 10^{23}kg)}{4.0\cdot 10^6 m}=-8.8\cdot 10^7 J

So now we can use eq.(1) to find the final kinetic energy:

K_f = K_i + U_i - U_f = 5.0\cdot 10^7 J+(-1.07\cdot 10^8 J)-(-8.8\cdot 10^7 J)=3.1\cdot 10^7 J

(b) 6.3\cdot 10^7 J

The probe reaches a maximum distance of

r=8.0\cdot 10^6 m

which means that at that point, the kinetic energy is zero: (the probe speed has become zero):

K_f = 0

At that point, the gravitational potential energy is

U_f=-G\frac{mM}{r}=-(6.67\cdot 10^{-11})\frac{(10 kg)(5.3\cdot 10^{23}kg)}{8.0\cdot 10^6 m}=-4.4\cdot 10^7 J

So now we can use eq.(1) to find the initial kinetic energy:

K_i = K_f + U_f - U_i = 0+(-4.4\cdot 10^7 J)-(-1.07\cdot 10^8 J)=6.3\cdot 10^7 J

3 0
2 years ago
What is the net force of a 25 g object with an acceleration of 3 m/s^2? 100PTS
gayaneshka [121]

Explanation:

Force=Mass×acceleration

force=25×3

force=75N

5 0
2 years ago
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Of gold, oxygen, iron, and sulfur, which is a renewable source?
Ulleksa [173]
Oxygen bc plants go through photosynthesis which keeps producing more oxygen
6 0
3 years ago
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