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2 years ago

Eva Baul throws a ball upward at 23.4 m/s

Physics

1 answer:

Mice21 [21]2 years ago

6 0

Answer:

28 m

Explanation:

v₀ = 23.4 m/s, g = -9.8 m/s²

at the peak v = 0

find h

v² - v₀² = 2gh

0 - 23.4² = 2(-9.8)h = -19.6 h

so h = 28 m

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If the mass of a planet is 0.231 mE and its radius is 0.528 rE, estimate the gravitational field g at the surface of the planet.

crimeas [40]

Answer:

$8.1 m/s^2$

Explanation:

The strength of the gravitational field at the surface of a planet is given by

$g=\frac{GM}{R^2}$ (1)

where

G is the gravitational constant

M is the mass of the planet

R is the radius of the planet

For the Earth:

$g_E = \frac{GM_E}{R_E^2}=9.8 m/s^2$

For the unknown planet,

$M_X = 0.231 M_E\\R_X = 0.528 R_E$

Substituting into the eq.(1), we find the gravitational acceleration of planet X relative to that of the Earth:

$g_X = \frac{GM_X}{R_X^2}=\frac{G(0.231M_E)}{(0.528R_E)^2}=\frac{0.231}{0.528^2}(\frac{GM_E}{R_E^2})=0.829 g_E$

And substituting g = 9.8 m/s^2,

$g_X = 0.829(9.8)=8.1 m/s^2$

3 0

3 years ago

The action of two forces. One is a forward force of 1157 N provided by traction between the wheels and the road. The other is a

Pie

Complete question

A 2700 kg car accelerates from rest under the action of two forces. one is a forward force of 1157 newtons provided by traction between the wheels and the road. the other is a 902 newton resistive force due to various frictional forces. how far must the car travel for its speed to reach 3.6 meters per second? answer in units of meters.

Answer:

The car must travel 68.94 meters.

Explanation:

First, we are going to find the acceleration of the car using Newton's second Law:

$\sum\overrightarrow{F}=m\overrightarrow{a}$ (1)

with m the mass , a the acceleration and $\sum\overrightarrow{F}$ the net force forces that is:

$(F-f)$ (2)

with F the force provided by traction and f the resistive force:

(2) on (1):

$(F-f)=ma$

solving for a:

$a=\frac{F-f}{m} =\frac{1157N-902N}{2700kg} =0.094\frac{m}{s^{2}}$

Now let's use the Galileo’s kinematic equation

$Vf^{2}=Vo^{2}+2a\varDelta x$ (3)

With Vo te initial velocity that's zero because it started from rest, Vf the final velocity (3.6) and $\varDelta x$ the time took to achieve that velocity, solving (3) for $\varDelta x$ :