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Vikentia [17]
3 years ago
9

A robot probe drops a camera off the rim of a 278 m high cliff on Mars, where the free-fall acceleration is 3.7 m/s2 . Find the

velocity with which it hits the ground. Answer in units of m/s.
Physics
1 answer:
FromTheMoon [43]3 years ago
4 0
S = u + at u = 0 278 = 3.7t t = 278/3.7 = 75.135.. v = ut + 0.5at^2 u = 0 v = 0.5 * 3.7 * 75.135^2 = 10,443 m/sec
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3. The direction of the centripetal force acting on a ball on a string being spun in a circle around a person is
ruslelena [56]

D. center seeking. Good Luck

4 0
3 years ago
Read 2 more answers
Questions<br> 2.<br> Rewrite the following quantities using suitable<br> prefixes<br><br> 5000 000
Rufina [12.5K]
Ans. 5 X 10^6

Explanation:

Here,^ represents 6 times 10 and 5 is multiplied. That is 5000 000.

Hope this helps
5 0
3 years ago
While accelerating at 5.22 m/s/s an object changes its velocity from 6.73 m/s to 29.88 m/s. Over what
yan [13]

Answer:

Explanation:

v² = u² + 2as

s = (v² - u²) / 2a

s = (29.88² - 6.73²) / (2(5.22))

s = 81.1802203065...

s = 81.18 m

4 0
3 years ago
It you balance a weight of 20 N at the 15 cm MARK and want to balance a weight of on the other side of the meter stick at 70 cm
agasfer [191]

Answer:

The weight that will balance the meter stick at 70 cm mark is 35N

Explanation:

Given;

first weight at 15 cm mark, W₁ = 20 N

second weight at 70 cm mark, W₂ = ?

A sketch of the question in diagram form;

The center of gravity of the meter stick is 50 cm

                     15cm               50cm         70cm

0---------------------------------------Δ------------------------------------------100cm

                     ↓                                           ↓

                     20N                                      W₂

                     <------------------>|<--------------->

                                35cm               20cm

Take moment about the pivot;

Clockwise moment = anticlockwise moment

W₂(20cm) = 20N(35 cm)

W₂ = (20 x 35) / (20)

W₂ = 35 N

Therefore, the weight that will balance the meter stick at 70 cm mark is 35N

8 0
3 years ago
A 4500 kg car is pushed by 3 boys who exert a force totaling 1500 N. When the boys begin to push the car it is traveling 4.5 m/s
Brilliant_brown [7]

Answer:

The velocity of the car after 3 seconds is 5.5 m/s

Explanation:

Mathematically, we can get the acceleration of the car first;

F = m•a

where m is the mass, F is the applied force and a is the acceleration

from the question;

F = 1500 N , m = 4500 kg and a = ?

By rewriting the formula;

a = F/m

a = 1500/4500 = 1/3 m/s^2

Now to get the velocity of the car after 3 seconds;

Mathematically;

V = U + at

where V is the final

velocity, U is the initial velocity,

From the question, we are trying to get V;

where U = 4.5 m/s

a = 1/3 m/s^2 and t = 3 seconds

Thus;

V = 4.5 + (1/3)(3)

V = 4.5 + 1

V = 5.5 m/s

7 0
2 years ago
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