All categories

3 years ago

A force of 100 newtons was necessary to lift a rock. a total of 150 joules of work was done. how far was the rock lifted

Physics

1 answer:

Makovka662 [10]3 years ago

4 0

Answer:

1.5m

Explanation:

150÷100=1.5

Work done ÷force

You might be interested in

PLZ HELP I DONT GET IT

GaryK [48]

Answer:

0.1 L

Explanation:

From the question given above, we obtained the following data:

Initial volume (V₁) = 0.05 L

Initial Pressure (P₁) = 207 KPa

Final pressure (P₂) = 101 KPa

Final volume (V₂) =?

We can obtain the new volume (i.e the final volume) of the gas by using the Boyle's law equation as illustrated below:

P₁V₁ = P₂V₂

207 × 0.05 = 101 × V₂

10.35 = 101 × V₂

Divide both side by 101

V₂ = 10.35 / 101

V₂ = 0.1 L

Thus, the new volume of the gas is 0.1 L

6 0

3 years ago

Can someone explain it with steps?

Anton [14]

Answer:

Option C is the correct answer

Explanation:

Distance travelled by car during reaction time

$=15\times0.4\\\\=6m$

The car stopped before hitting the animal by $1 m$

Distance travelled during deceleration is $21-6-1=14m$

Hence by $v^2=u^2+2as$

We have

$0^2=15^2+2 \cdot a \cdot 14\\\\a=\frac{-225}{28} \\\\=-8.03m/s^2$

Option C is the correct answer

5 0

3 years ago

Read 2 more answers

If you drop an object from rest, the distance it falls is given by (1/2)at2, where a is the acceleration of the object and t is

klio [65]

If,

$d \: = \: \frac{1}{2} a {t}^{2}$

then, with 3x time t, (suppose, new distance is h)

$h \: = \: \frac{1}{2} a {(3t)}^{2}$

$= \frac{1}{2} a9 {t}^{2}$

$= 9 \: \frac{1}{2} a{t}^{2}$

$= 9d$

Therefore, new distance h will be 9 times bigger than distance d.

answer: c

7 0

3 years ago

The river narrows at a rapids from a width of 12 m to a width of only 5.8 m. The depth of the river before the rapids is 2.7 m;

Alisiya [41]

Answer:

7.89 m/s

Explanation:

Given that

Width of the river, b1 = 12 m

Width of the river, b2 = 5.8 m

Depth of the river, d1 = 2.7 m

Depth of the river, d2 = 0.85 m

Speed of the river, v1 = 1.2 m/s

Speed of the river, v2 = ?

Area of the river before the rapid, a1 = 12 * 2.7 = 32.4 m²

Area of the river after the rapid, a2 = 5.8 * 0.85 = 4.93 m²

To solve this question, we use a relation between the speed of the river and the volume of the river. We say,

Area1 * velocity1 = Area2 * velocity2, and when we substitute the values for each other we have

32.4 * 1.2 = 4.93 * v2

38.88 = 4.93v2

v2 = 38.88 / 4.93

v2 = 7.89 m/s

Therefore, the speed of the river after the rapid is 7.89 m/s

6 0

3 years ago

A 99.1-kg baseball player slides into second base. The coefficient of kinetic friction between the player and the ground is μk =

Stels [109]

Answer:

628.022466 N

8.61 m/s

Explanation:

m = Mass

$\mu$ = Coefficient of friction

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

g = Acceleration due to gravity = 9.81 m/s²

$F_f=\mu mg\\\Rightarrow F_f=0.646\times 99.1\times 9.81\\\Rightarrow F_f=628.022466\ N$

Magnitude of frictional force is 628.022466 N

$F=ma\\\Rightarrow a=\frac{F_f}{m}\\\Rightarrow a=\frac{628.022466}{99.1}\\\Rightarrow a=6.33726\ m/s^2$

$v=u+at\\\Rightarrow 0=u-6.33726\times 1.36\\\Rightarrow u=8.61\ m/s$

Initial speed of the player is 8.61 m/s

4 0

3 years ago