All categories

2 years ago

Some expressways are curved, banked and designed to maximize ___________ at higher speeds.

Physics

1 answer:

Liono4ka [1.6K]2 years ago

6 0

Answer:

Safety

Explanation:

Expressways are banked to resist centifugal action

You might be interested in

The voltage in a circuit can be determined from the ____, total resistance, and capacitance

Dmitrij [34]

I think the answer is D, but I’m not at all sure :l

4 0

3 years ago

Consider four point charges arranged in a square with sides of length L. Three of the point charges have charge q and one of the

nydimaria [60]

Answer: $F_{net}=\frac{kq^2}{(L)^2}\left [ \frac{1}{2}+\sqrt{2}\right ]$

Explanation:

Given

Three charges of magnitude q is placed at three corners and fourth charge is placed at last corner with -q charge

Force due to the charge placed at diagonally opposite end on -q charge

$F_1=\frac{kq(-q)}{(L\sqrt{2})^2}$

where $L\sqrt{2}=$ Distance between the two charges

$F_1=-\frac{kq^2}{2L^2}$

negative sign indicates that it is an attraction force

Now remaining two charges will apply the same amount of force as they are equally spaced from -q charge

$F_2=\frac{kq(-q)}{(L)^2}$

The magnitude of force by both the charge is same but at an angle of $90^{\circ}$

thus combination of two forces at 2 and 3 will be

$F'=\sqrt{2}\frac{kq^2}{2L^2}$

Now it will add with force due to 1 charge

Thus net force will be

$F_{net}=\frac{kq^2}{(L)^2}\left [ \frac{1}{2}+\sqrt{2}\right ]$

6 0

2 years ago

How much force is required to move a electron through an electric field with strength of 1.375 x 10^19 N/C?

Luda [366]

Answer:

The magnitude of the force required to move the electron through the given field is 2.203 N

Explanation:

Given;

The field strength of the electron, E = 1.375 x 10¹⁹ N/C

charge of electron, q = 1.602 x 10⁻¹⁹ C

The magnitude of the force required to move the electron through the given field is calculated as follows;

F = Eq

F = (1.375 x 10¹⁹ N/C) (1.602 x 10⁻¹⁹ C)

F = 2.203 N

Therefore, the magnitude of the force required to move the electron through the given field is 2.203 N

4 0

3 years ago

on a very muddy football field, a 120 kg linebacker tackles an 75 kg halfback. immediately before the collision, the linebacker

Aleksandr-060686 [28]

B4 the tackle:

The linebacker's momentum = 115 x 8.5 = 977.5 kg m/s north 

and the halfback's momentum = 89 x 6.7 = 596.3 kg m/s east 

After the tackle they move together with a momentum equal to the vector sum of their separate momentums b4 the tackle 

The vector triangle is right angled: 

magnitude of final momentum = √(977.5² + 596.3²) = 1145.034 kg m/s 

so (115 + 89)v(f) = 1145.034 ←←[b/c p = mv] 

v(f) = 5.6 m/s (to 2 sig figs) 

direction of v(f) is the same as the direction of the final momentum 

so direction of v(f) = arctan (596.3 / 977.5) = N 31° E (to 2 sig figs) 

so the velocity of the two players after the tackle is 5.6 m/s in the direction N 31° E 

btw ... The direction can be given heaps of different ways ... N 31° E is probably the easiest way to express it when using the vector triangle to find it

4 0

3 years ago

When hiking at altitude, the air pressure drops to 825,000Pa. Solve the force that site exerts against the side of a rectangular

OLga [1]

Answer:

The force is $1237500N$