3 years ago

Which force is a field force?

Physics

2 answers:

jok3333 [9.3K]3 years ago

7 0

it has to me c because it’s a part of the force of things that do

koban [17]3 years ago

6 0

Electric force

sxjxkxkxkxkxkxk

You might be interested in

Which factor is most important in determining climate?

alisha [4.7K]

Temperature. The other three dont have anything to do with determining climate

8 0

3 years ago

Read 2 more answers

An artillery shell is launched on a flat, horizontal field at an angle of α = 40.8° with respect to the horizontal and with an i

Lina20 [59]

Answer:

1317.4 m

Explanation:

We are given that

Angle= $\alpha=40.8^{\circ}$

Initial speed = $v_0=346m/s$

We have to find the horizontal distance covered by the shell after 5.03 s.

Horizontal component of initial speed= $v_{ox}=v_0cos\theta=346cos40.8=261.9m/s$

Vertical component of initial speed= $v_{oy}=346sin40.8=226.1m/s$

Time=t=5.03 s

Horizontal distance = $Horizontal\;velocity\times time$

Using the formula

Horizontal distance= $261.9\times 5.03$

Horizontal distance=1317.4 m

Hence, the horizontal distance covered by the shell=1317.4 m

8 0

3 years ago

Janice has just measured the density of an object. Which value is possible? (Density: D = )

babunello [35]

It is 6 g/cm3 because density cannot be negative, and it is not speed in which the unit would be m/s.

6 0

3 years ago

Read 2 more answers

What is the maximum value the string tension can have before the can slips? The coefficient of static friction between the can a

Naya [18.7K]

Answer:

T= 38.38 N

Explanation:

Here

mass of can = m = 3 kg

g= 9.8 m/sec2

angle θ = 40°

From figure we see the vertical and horizontal component of tension force T

If the can is to slip - then horizontal component of tension force should become equal to force of friction.

First we find force of friction

Fs= μ R

where

μ = 0.76

R = weight of can = mg = 3 × 9.8 = 29.4 N

Now horizontal component of tension

Tx= T cos 40 = T× 0.7660 N

==>T× 0.7660 = 29.4

==> T= 38.38 N

8 0

3 years ago

34. A train, starting from rest, accelerates along the platform at a uniform rate of 0.6 m/s2. A passenger standing on the platf

ira [324]

Answer:

4.08 s

Explanation:

Let the passenger took "t" time to catch the train

so in this case the total distance moved by the train + 5 m = total distance moved by the passenger

so we will have

distance moved by train is given as

$d_1 = \frac{1}{2}(0.6) t^2$

also the distance moved by passenger

$d_2 = \frac{1}{2}(1.2) t^2$

so we will have

$d_1 + 5 = d_2$

$0.3 t^2 + 5 = 0.6 t^2$

$0.3 t^2 = 5$

t = 4.08 s

3 0

3 years ago