Answer:
Approximately 
, assuming that 
.
Explanation:
Let 
and 
denote the mass and acceleration of Spiderman, respectively.
There are two forces on Spiderman:
- Downward gravitational attraction from the earth:
. - Upward tension force from the strand of web
.
The directions of these two forces are exactly opposite of one another. Besides, because Spiderman is accelerating upwards, the magnitude of (which points upwards) should be greater than that of 
(which points downwards towards the ground.)
Subtract the smaller force from the larger one to find the net force on Spiderman:
.
On the other hand, apply Newton's Second Law of motion to find the value of the net force on Spiderman:
.
Combine these two equations to get:
.
Therefore:
.
By Newton's Third Law of motion, Spiderman would exert a force of the same size on the strand of web. Hence, the size of the force in the strand of the web should be approximately 
(downwards.)
In will most likely decrease its speed.
hope this helps.
Answer:
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Explanation:
Answer:
The automobile's acceleration in that time interval is -2 m/s^2
Explanation:
The acceleration is defined as the rate of change of the velocity.
The average acceleration in a given lapse of time is calculated as:
A = (final velocity - initial velocity)/time.
In this case, we have:
initial velocity = 31 m/s
final velocity = 15 m/s
time = 8 seconds.
Then the average acceleration is:
A = (15m/s - 31m/s)/8s = -2 m/s^2
Answer:
i)-6.25m/s
ii)18 metres
iii)26.5 m/s or 95.4 km/hr
Explanation:
Firstly convert 90km/hr to m/s
90 × 1000/3600 = 25m/s
(i) Apply v^2 = u^2 + 2As...where v(0m/s) is the final speed and u(25m/s) is initial speed and also s is the distance moved through(50 metres)
0 = (25)^2 + 2A(50)
0 = 625 + 100A....then moved the other value to one
-625 = 100A
Hence A = -6.25m/s^2(where the negative just tells us that its deceleration)
(ii) Firstly convert 54km/hr to m/s
In which this is 54 × 1000/3600 = 15m/s
then apply the same formula as that in (i)
0 = (15)^2 + 2(-6.25)s
-225 = -12.5s
Hence the stopping distance = 18metres
(iii) Apply the same formula and always remember that the deceleration values is the same throughout this question
0 = u^2 + 2(-6.25)(56)
u^2 = 700
Hence the speed that the car was travelling at is the,square root of 700 = 26.5m/s
In km/hr....26.5 × 3600/1000 = 95.4 km/hr
