An object moving at 30 m/s takes 5 sec. to come to a stop. What is the object’s acceleration

A raised plaster cover (often called a plaster ring or mud ring) is permitted to increase the maximum number of conductors permi

andreev551 [17]

Answer:

When it is marked with its cubic-inch volume

Explanation:

Because this allows for best and efficient identification

6 0

2 years ago

Suppose that the electric field in the Earth's atmosphere is E = 8.60 101 N/C, pointing downward. Determine the electric charge

asambeis [7]

Answer:

q = 3.87 x 10⁵ C

Explanation:

given,

Electric field, E = 8.60 x 10¹ = 86 N/C

radius of earth, R = 6371 Km = 6.371 x 10⁶ m

Coulomb constant, K = 9 x 10⁹ N · m²/C²

Charge on the earth = ?

the electric field at the point

$E =\dfrac{kq}{r^2}$

$q =\dfrac{Er^2}{k}$

inserting all the values

$q =\dfrac{86\times (6.371\times 10^6)^2}{9\times 10^{9}}$

q = 3.87 x 10⁵ C

The electric charge on the earth is equal to 3.87 x 10⁵ C

4 0

3 years ago

please help. Sound travels at 330 m/s. If a lightning bolt strikes the ground 5 m away from you, how long will it take for the s

Olenka [21]

Answer:

0.015 seconds

Explanation:

5/330=0.015

3 0

2 years ago

1.)Two objects, one of m=20,000 kg, and another of 12,500 kg, are placed at a distance of 5 meters apart. What is the force of g

Delvig [45]

1) $6.67\cdot 10^{-4} N$

The force of gravitation between the two objects is given by:

$F=G\frac{m_1 m_2}{r^2}$

where

$G=6.67\cdot 10^{-11} kg^{-1} m^{3} s^{-2}$ is the gravitational constant

m1 = 20,000 kg is the mass of the first object

m2 = 12,500 kg is the mass of the second object

r = 5 m is the distance between the two objects

Substituting the numbers inside the equation, we find

$F=(6.67\cdot 10^{-11})\frac{(20,000 kg)(12,500 kg)}{(5 m)^2}=6.67\cdot 10^{-4} N$

2) $2.7\cdot 10^{-3} N$

From the formula in exercise 1), we see that the force is inversely proportional to the square of the distance:

$F \sim \frac{1}{r^2}$

this means that if we cut in a half the distance without changing the masses, the magnitude of the forces changes by a factor

$F'\sim \frac{1}{(r/2)^2}=4 \frac{1}{r^2}=4F$

So, the gravitational force increases by a factor 4. Therefore, the new force will be

$F' = 4 F=4(6.67\cdot 10^{-4} N)=2.7\cdot 10^{-3} N$

3) 12.5 Nm

The torque is equal to the product between the magnitude of the perpendicular force and the distance between the point of application of the force and the centre of rotation:

$\tau=Fd$

Where, in this case:

F = 25 N is the perpendicular force

d = 0.5 m is the distance between the force and the center

By using the equation, we find

$\tau=(25 N)(0.5 m)=12.5 Nm$

4) 0.049 kg m^2/s

The relationship between angular momentum (L), moment of inertia (I) and angular velocity ( $\omega$ ) is:

$L=I\omega$

In this problem, we have

$I=0.007875 kgm^2$

$\omega=6.28 rad/s$

So, the angular momentum is

$L=I\omega=(0.007875 kgm^2)(6.28 rad/s)=0.049 kg m^2/s$

6 0

2 years ago

Please help me with this question guys.

katen-ka-za [31]

Answer:

<em>The average speed is 22.2 km/h</em>

Explanation:

<u>Average Speed</u>

Given an object travels a total distance d and took a total time t, then the average speed is:

$\displaystyle \bar v=\frac{d}{t}$

The mailman first drives d1=7 km at v1=15 km/h. The time taken to drive is:

$\displaystyle t1=\frac{d1}{v1}=\frac{7}{15}=0.467\ h$

Then he drives d2=7 km at v2=43 km/h taking a time of:

$\displaystyle t2=\frac{d2}{v2}=\frac{7}{43}=0.163\ h$

The total time is

t=0.467 h + 0.163 h = 0.63 h

The total distance is

d = 7 km + 7 km = 14 km

The average speed is:

$\displaystyle \bar v=\frac{14}{0.63}=22.2\ km/h$

The average speed is 22.2 km/h

7 0

2 years ago

An object moving at 30 m/s takes 5 sec. to come to a stop. What is the object’s acceleration​

An object moving at 30 m/s takes 5 sec. to come to a stop. What is the object’s acceleration