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d1i1m1o1n [39]
3 years ago
7

Determine the mass of a boat that is propelled forward by a force of 1250N at an acceleration of 3 m/s2.

Physics
1 answer:
Pavlova-9 [17]3 years ago
6 0

Answer:

416.6 KG

Explanation:

Due to the mass being missing from the equation you divide the force by the acceleration

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Standing at a crosswalk, you hear a frequency of 550 Hz from the siren of an approaching ambulance. After the ambulance passes,
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There are six steps to this process , I uploaded step one and as you can see you can get all six on Quizlet:). Good luck

6 0
3 years ago
A small apple weighs 0.8N. How much would an apple that is 4 times as massive weigh?
-BARSIC- [3]

Answer:

W=31.99 N

Explanation:

W= 8 N (N= kg*m/ s^2)

W= m* g

8 N= m * 9.8 m/s^2

m= 0.816 kg

m*4= (0.816)4

= 3.264 kg

W= mg

= (3.264 kg)(9.8 m/s^2)

W=31.99 N

4 0
3 years ago
The acceleration due to gravity on the surface of Mars is about one-third the acceleration due to gravity on Earth’s surface.
aksik [14]

Answer:

one-third of its weight on Earth's surface

Explanation:

Weight of an object is = W = m*g

Gravity on Earth = g₁ = 9.8 m/s

Gravity on Mars = g₂ = \frac{1}{3} g₁

Weight of probe on earth = w₁ = m * g₁

Weight of probe on Mars = w₂ = m * g₂ -------- ( 1 )

As g₂ = g₁/3 --------- ( 2 )

Put equation (2) in equation (1)

so

Weight of probe on Mars = w₂ = m * g₁ /3

Weight of probe on Mars = \frac{1}{3}  m * g₁ = \frac{1}{3} w₁

⇒Weight of probe on Mars =\frac{1}{3} Weight of probe on earth

6 0
4 years ago
It took David 2 seconds to lift a 5 Newton bag of toys from the floor to the top of the top shelf, which is 3 meters tall. How m
just olya [345]
So power is equal to work over time and work is force times distance, you do 5 times 3 and get 15 dividing by 2 gives us 7.5 W answer c
7 0
3 years ago
Two points charges are brought closer together,increasing the force between them by a factor of 25.By what factor wa their separ
Roman55 [17]

Answer:

The separation between the charges was decreased by a factor of 0.2

Explanation:

The Coulomb's force between two charges is given by;

F = \frac{kq^2}{r^2} \\\\let \ kq^2 \ be \ constant\\\\F_1r_1^2 = F_2r_2^2\\\\r_2^2 = \frac{F_1r_1^2}{F_2} \\\\increasing \ the \ force \ between \ them \ by \ factor \ of \ 25\\\\(F_2 = 25F_1)\\\\r_2^2 = \frac{F_1r_1^2}{25F_1}\\\\r_2^2 = \frac{r_1^2}{25}\\\\r_2 = \sqrt{\frac{r_1^2}{25} }\\\\ r_2 = \frac{r_1}{5}

r₂ = 0.2r₁

Therefore, the separation between the charges was decreased by a factor of 0.2.

6 0
3 years ago
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