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Alborosie
3 years ago
8

1,021 kg x 9.8m/s/s =

Physics
2 answers:
drek231 [11]3 years ago
3 0
The anwser is 10005.8 newtons
Sergeu [11.5K]3 years ago
3 0
<span>10 005.8 newtons is your answer, hope this helps!</span>
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 Why does a child in a wagon seem to fall backward when you give the wagon a sharp pull forward? 15. What force is needed to acc
slavikrds [6]

Answer:

Force, F = 77 N

Explanation:

A child in a wagon seem to fall backward when you give the wagon a sharp pull forward. It is due to Newton's third law of motion. The forward pull on wagon is called action force and the backward force is called reaction force. These two forces are equal in magnitude but they acts in opposite direction.

We need to calculate the force is needed to accelerate a sled. It can be calculated using the formula as :

F = m × a

Where

m = mass = 55 kg

a = acceleration = 1.4 m/s²

F=55\ kg\times 1.4\ m/s^2

F = 77 N

So, the force needed to accelerate a sled is 77 N. Hence, this is the required solution.

3 0
3 years ago
Emmett is lifting a box vertically. Which forces are necessary for calculating the total force?
GrogVix [38]

When Emmett is lifting a box vertically, the forces that must be added to calculate the total force are: the gravitational force, tension force(the force exerted by Emmett to the box and the force exerted by the box to Emmett), and air resistance force.

4 0
3 years ago
How much mass should be attached to a vertical ideal spring having a spring constant (force constant) of 39.5 n/m so that it wil
mrs_skeptik [129]
The frequency of a simple harmonic oscillator such as a spring-mass system is given by
f= \frac{1}{2 \pi}   \sqrt{ \frac{k}{m} }
where 
k is the spring constant
m is the mass attached to the spring.

Re-arranging the formula, we get:
m= \frac{k}{4 \pi^2 f^2}
and since we know the constant of the spring:
k=39.5 N/m
and the frequency of oscillation:
f=1.00 Hz
we can find the value of the mass attached to it:
m= \frac{39.5 Hz}{4 \pi^2 (1.00 Hz)^2} = 1.00 kg
7 0
3 years ago
An 8.0 cm object is 40.0 cm from a concave mirror that has a focal length of 10.0 cm. Its image is 16.0 cm in front of the mirro
svet-max [94.6K]
 We can rearrange the mirror equation before plugging our values in. 
1/p = 1/f - 1/q. 
1/p = 1/10cm - 1/40cm
1/p = 4/40cm - 1/40cm = 3/40cm
40cm=3p  <-- cross multiplication
13.33cm = p

Now that we have the value of p, we can plug it into the magnification equation.

M=-16/13.33=1.2
1.2=h'/8cm
9.6=h'

So the height of the image produced by the mirror is 9.6cm.
6 0
3 years ago
At a particular instant, a proton at the origin has velocity &lt; 5e4, -2e4, 0&gt; m/s. You need to calculate the magnetic field
vesna_86 [32]

Answer:

9.7\times 10^{-5} T

Explanation:

Velocity =5\times 10^4i-2\times 10^4j

r=0.03i+0.05j

r=\mid r\mid=\sqrt{(0.03)^2+(0.05)^2}=0.058

v=\mid V\mid=\sqrt{(5\times 10^4)^2+(-2\times 10^{4})^2}=5.39\times 10^{2}

We know that

B=\frac{mv}{qr}

Where q=1.6\times 10^{-19} C

Mass of proton=1.67\times 10^{-27} kg

Using the formula

B=\frac{1.67\times 10^{-27}\times 5.39\times 10^2}{1.6\times 10^{-19}\times 0.058}

B=9.7\times 10^{-5} T

3 0
3 years ago
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