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2 years ago

15

Why does the arrow labeled F norm point upward?

2 answers:

ohaa [14]2 years ago

8 0

The answer is D I think

jenyasd209 [6]2 years ago

5 0

I think the answer is d too

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A cheetah can run at a maximum speed 103 km/h and a gazelle can run at a maximum speed of 76.5 km/h. If both animals are running

Katena32 [7]

Answer:

1

$t_a = 13.49 \ s$

2

The distance is $D = 55.2 \ m$

Explanation:

From the question we are told that

The maximum speed of the cheetah is $v = 103 \ km/h = 28.61 \ m/s$

The maximum of gazelle is $u = 76.5 \ km/h = 21.25 \ m/s$

The distance ahead is $d = 99.3 \ m$

Let $t_a$ denote the time which the cheetah catches the gazelle

Gnerally the equation representing the distance the cheetah needs to move in order to catch the gazelle is

$v* t_a = d + u* t_a$

=> $28.61 t_a = 99.3 + 21.25t_a$

=> $7.36 t_a = 99.3$

=> $t_a = 13.49 \ s$

Now at t = 7.5 s

$7.5 v = D+ 7.5u$

=> $28.61 * 7.5 = D + 21.25* 7.5$

=> $7.36 * 7.5 =D$

=> $D = 55.2 \ m$

Hence the for the gazelle to escape the cheetah it must be 55.2 m

5 0

3 years ago

A truck skids for a distance of 25 m with the road pushing on its tires with force of 1500 N as its brakes

MakcuM [25]

Answer:-6800J

Explanation: 8.0m x 850N = 6800

Rewritten as a negative when brake/stop -6800

6 0

3 years ago

Read 2 more answers

A baseball hit just above the ground leaves the bat 27 m/s at 45° above the horizontal. A) How far away does the ball strike the

Sedbober [7]

Answer:

A) The ball hits the ground 74.45 m far from the hitting position.

B) Maximum height of the ball = 18.57 m

Explanation:

There are two types of motion in this horizontal and vertical motion.

We have velocity = 27 m/s at 45° above the horizontal

Horizontal velocity = 27cos45 = 19.09 m/s

Vertical velocity = 27sin45 = 19.09 m/s

Time to reach maximum height,

v = u + at

0 = 19.09 - 9.81 t

t = 1.95 s

So total time of flight = 2 x 1.95 = 3.90 s

A) So the ball travels at 19.09 m/s for 3.90 seconds.

Horizontal distance traveled = 19.09 x 3.90 = 74.45 m

So the ball hits the ground 74.45 m far from the hitting position.

B) We have vertical displacement

S = ut + 0.5 at²

H = 19.09 x 1.95 - 0.5 x 9.81 x 1.95² = 18.57 m

Maximum height of the ball = 18.57 m

6 0

3 years ago

An example of a folkway violation would be driving 70 mph in a 50-mph zone. Group of answer choices True False

shutvik [7]

Answer:

Yes.

Explanation:

That is also a law violation.

3 0

3 years ago

In an attempt to reduce the extraordinarily long travel times for voyaging to distant stars, some people have suggested travelin

alexandr402 [8]

Answer:

a) $v=0.999124c$

b) $E=7.566*10^{22}$

c) $E_a=760 times\ larger$

Explanation:

From the question we are told that

Distance to Betelgeuse $d_b=430ly$

Mass of Rocket $M_r=20000$

Total Time in years traveled $T_d=36years$

Total energy used by the United States in the year 2000 $E_{2000}=1.0*10^20$

Generally the equation of speed of rocket v mathematically given by

$v=\frac{2d}{\triangle t}$

$v=860ly/ \triangle t$

where

$\triangle t=\frac{\triangle t'}{(\sqrt{1-860/ \triangle t)^2}}$

$\triangle t=\frac{36}{(\sqrt{1-860/ \triangle t)^2}}$

$\triangle t=\sqrt{(860)^2+(36)^2}$

$\triangle t=860.7532$

Therefore

$v=\frac{860ly}{ 860.7532}$

$v=0.999124c$

b)

Generally the equation of the energy E required to attain prior speed mathematically given by

$E=\frac{1}{\sqrt{1-(v/c)^2} }-1(20000kg)(3*10^8m/s)^2$

$E=7.566*10^{22}$

c)Generally the equation of the energy $E_a$ required to accelerate the rocket mathematically given by

$E_a=\frac{E}{E_{2000}}$

$E_a=\frac{7.566*10^{22}}{1.0*10^{20}}$

$E_a=760 times\ larger$

8 0

3 years ago